Học tại trường Chưa có thông tin
Đến từ Thành phố Hồ Chí Minh , Chưa có thông tin
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Điểm SP 115878

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Câu trả lời:

1: \(24=2^3\cdot3;30=2\cdot3\cdot5\)

=>\(ƯCLN\left(24;30\right)=2\cdot3=6\)

=>ƯC(24;30)=Ư(6)={1;-1;2;-2;3;-3;6;-6}

2: \(12=2^2\cdot3;18=2\cdot3^2\)

=>\(ƯCLN\left(12;18\right)=2\cdot3=6\)

=>ƯC(12;18)=Ư(6)={1;-1;2;-2;3;-3;6;-6}

3: \(12=2^2\cdot3;20=2^2\cdot5\)

=>\(ƯCLN\left(12;20\right)=2^2=4\)

=>ƯC(12;20)=Ư(4)={1;-1;2;-2;4;-4}

4: \(15=3\cdot5;20=2^2\cdot5\)

=>ƯCLN(15;20)=5

=>ƯC(15;20)=Ư(5)={1;-1;5;-5}

5: \(20=2^2\cdot5;30=2\cdot3\cdot5\)

=>\(ƯCLN\left(20;30\right)=2\cdot5=10\)

=>ƯC(20;30)=Ư(10)={1;-1;2;-2;5;-5;10;-10}

6: \(12=2^2\cdot3;36=2^2\cdot3^2\)

=>\(ƯCLN\left(12;36\right)=2^2\cdot3=12\)

=>ƯC(12;36)=Ư(12)={1;-1;2;-2;3;-3;4;-4;6;-6;12;-12}

7: \(18=2\cdot3^2;30=2\cdot3\cdot5\)

=>\(ƯCLN\left(18;30\right)=2\cdot3=6\)

=>ƯC(18;30)=Ư(6)={1;-1;2;-2;3;-3;6;-6}

8: \(24=2^3\cdot3;40=2^3\cdot5\)

=>\(ƯCLN\left(24;40\right)=2^3=8\)

=>ƯC(24;40)=Ư(8)={1;-1;2;-2;4;-4;8;-8}

9: \(45=3^2\cdot5;60=2^2\cdot3\cdot5\)

=>\(ƯCLN\left(45;60\right)=3\cdot5=15\)

=>ƯC(45;60)=Ư(15)={1;-1;3;-3;5;-5;15;-15}

10: \(30=2\cdot3\cdot5;40=2^3\cdot5\)

=>\(ƯCLN\left(30;40\right)=2\cdot5=10\)

=>ƯC(30;40)=Ư(10)={1;-1;2;-2;5;-5;10;-10}

11: \(36=2^2\cdot3^2;54=2\cdot3^3\)

=>\(ƯCLN\left(36;54\right)=2\cdot3^2=18\)

=>ƯC(36;54)=Ư(18)={1;-1;2;-2;3;-3;6;-6;9;-9;18;-18}

12: \(40=2^3\cdot5;60=2^2\cdot5\cdot3\)

=>\(ƯCLN\left(40;60\right)=2^2\cdot5=20\)

=>ƯC(40;60)=Ư(20)={1;-1;2;-2;4;-4;5;-5;10;-10;20;-20}

13: \(60=2^2\cdot3\cdot5;90=2\cdot3^2\cdot5\)

=>\(ƯCLN\left(60;90\right)=2\cdot3\cdot5=30\)

=>ƯC(60;90)=Ư(30)={1;-1;2;-2;3;-3;5;-5;6;-6;10;-10;15;-15;30;-30}

14: \(50=2\cdot5^2;70=2\cdot5\cdot7\)

=>\(ƯCLN\left(50;70\right)=2\cdot5=10\)

=>ƯC(50;70)=Ư(10)={1;-1;2;-2;5;-5;10;-10}

15: \(60=2^2\cdot3\cdot5;80=2^4\cdot5\)

=>\(ƯCLN\left(60;80\right)=2^2\cdot5=20\)

=>ƯC(60;80)=Ư(20)={1;-1;2;-2;4;-4;5;-5;10;-10;20;-20}

Câu trả lời:

b: ĐKXĐ: x<>0; y<>0

\(\left\{{}\begin{matrix}\dfrac{3}{x}-\dfrac{4}{y}=2\\\dfrac{4}{x}-\dfrac{5}{y}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{12}{x}-\dfrac{16}{y}=8\\\dfrac{12}{x}-\dfrac{15}{y}=9\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{12}{x}-\dfrac{16}{y}-\dfrac{12}{x}+\dfrac{15}{y}=8-9=-1\\\dfrac{3}{x}-\dfrac{4}{y}=2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-\dfrac{1}{y}=-1\\\dfrac{3}{x}=\dfrac{4}{y}+2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\\dfrac{3}{x}=4+2=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=\dfrac{1}{2}\end{matrix}\right.\left(nhận\right)\)

a: ĐKXĐ: x<>0; y<>0

\(\left\{{}\begin{matrix}\dfrac{3}{x}-\dfrac{1}{y}=7\\\dfrac{2}{x}+\dfrac{1}{y}=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{x}-\dfrac{1}{y}+\dfrac{2}{x}+\dfrac{1}{y}=7+8\\\dfrac{2}{x}+\dfrac{1}{y}=8\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{5}{x}=15\\\dfrac{1}{y}=8-\dfrac{2}{x}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}\\\dfrac{1}{y}=8-2:\dfrac{1}{3}=8-6=2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{1}{3}\\y=\dfrac{1}{2}\end{matrix}\right.\left(nhận\right)\)

e: ĐKXĐ: y<>2x; y<>-x

\(\left\{{}\begin{matrix}\dfrac{3}{2x-y}-\dfrac{6}{x+y}=-1\\\dfrac{1}{2x-y}-\dfrac{1}{x+y}=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{3}{2x-y}-\dfrac{6}{x+y}=-1\\\dfrac{3}{2x-y}-\dfrac{3}{x+y}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{2x-y}-\dfrac{6}{x+y}-\dfrac{3}{2x-y}+\dfrac{3}{x+y}=-1-0\\\dfrac{1}{2x-y}-\dfrac{1}{x+y}=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-\dfrac{3}{x+y}=-1\\2x-y=x+y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=3\\x-2y=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x+y-x+2y=3-0=3\\x+y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=2\end{matrix}\right.\left(nhận\right)\)

f: ĐKXĐ: x<>-1; y<>3

\(\left\{{}\begin{matrix}\dfrac{5x}{x+1}+\dfrac{y}{y-3}=27\\\dfrac{2x}{x+1}-\dfrac{3y}{y-3}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{15x}{x+1}+\dfrac{3y}{y-3}=81\\\dfrac{2x}{x+1}-\dfrac{3y}{y-3}=4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{15x}{x+1}+\dfrac{3y}{y-3}+\dfrac{2x}{x+1}-\dfrac{3y}{y-3}=81+4=85\\\dfrac{5x}{x+1}+\dfrac{y}{y-3}=27\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{17x}{x+1}=85\\\dfrac{y}{y-3}=27-\dfrac{5x}{x+1}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{x+1}=5\\\dfrac{y}{y-3}=27-5\cdot5=2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}5x+5=x\\2y-6=y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{4}\\y=6\end{matrix}\right.\left(nhận\right)\)

Câu trả lời:

Bài 1:

a: \(9x^2\cdot\left(2x-3\right)=0\)

=>\(x^2\left(2x-3\right)=0\)

=>\(\left[{}\begin{matrix}x^2=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\end{matrix}\right.\)

b: (x-1)(3x-6)=0

=>3(x-1)(x-2)=0

=>(x-1)(x-2)=0

=>\(\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

c: (x+2)(3-3x)=0

=>3(1-x)(x+2)=0

=>(1-x)(x+2)=0

=>\(\left[{}\begin{matrix}1-x=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\\\end{matrix}\right.\)

d: \(\left(\dfrac{2}{3}x+6\right)\left(8-2x\right)=0\)

=>\(\left[{}\begin{matrix}\dfrac{2}{3}x+6=0\\8-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{3}x=-6\\2x=8\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=-6:\dfrac{2}{3}=-9\\x=\dfrac{8}{2}=4\end{matrix}\right.\)

e: \(\left(4x+2\right)\left(x^2+1\right)=0\)

=>4x+2=0

=>4x=-2

=>\(x=-\dfrac{2}{4}=-\dfrac{1}{2}\)

f: \(\left(3x-4\right)\left(x+1\right)\left(2x-1\right)=0\)

=>\(\left[{}\begin{matrix}3x-4=0\\x+1=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=-1\\x=\dfrac{1}{2}\end{matrix}\right.\)

g: \(\left(3x-2\right)^2\cdot\left(x+1\right)\cdot\left(x-2\right)=0\)

=>\(\left[{}\begin{matrix}3x-2=0\\x+1=0\\x-2=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-1\\x=2\end{matrix}\right.\)

h: \(\left(2x+3\right)^2=\left(x-5\right)^2\)

=>\(\left(2x+3\right)^2-\left(x-5\right)^2=0\)

=>(2x+3+x-5)(2x+3-x+5)=0

=>(3x-2)(x+8)=0

=>\(\left[{}\begin{matrix}3x-2=0\\x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-8\end{matrix}\right.\)

i: \(\left(6x-7\right)\left(3x+4\right)=\left(7-6x\right)\left(x-1\right)\)

=>\(\left(6x-7\right)\left(3x+4\right)+\left(6x-7\right)\left(x-1\right)=0\)

=>(6x-7)(3x+4+x-1)=0

=>(6x-7)(4x+3)=0

=>\(\left[{}\begin{matrix}6x-7=0\\4x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{6}\\x=-\dfrac{3}{4}\end{matrix}\right.\)

j: \(\left(3x-2\right)\left(x+1\right)=x^2-1\)

=>(3x-2)(x+1)-(x-1)(x+1)=0

=>(x+1)(3x-2-x+1)=0

=>(x+1)(2x-1)=0

=>\(\left[{}\begin{matrix}x+1=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{2}\end{matrix}\right.\)

k: \(-5\left(4x-1\right)\left(x-2\right)=2\left(4x-1\right)^2\)

=>\(2\left(4x-1\right)^2+5\left(4x-1\right)\left(x-2\right)=0\)

=>\(\left(4x-1\right)\left[2\left(4x-1\right)+5\left(x-2\right)\right]=0\)

=>\(\left(4x-1\right)\left(8x-2+5x-10\right)=0\)

=>(4x-1)(13x-12)=0

=>\(\left[{}\begin{matrix}4x-1=0\\13x-12=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=\dfrac{12}{13}\end{matrix}\right.\)

l: \(x^2-8x+12=0\)

=>(x-2)(x-6)=0

=>\(\left[{}\begin{matrix}x-2=0\\x-6=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)

Bài 2:

a: ĐKXĐ: x<>0

\(\dfrac{1}{x}+\dfrac{1}{2x}=\dfrac{3}{2}\)

=>\(\dfrac{2}{2x}+\dfrac{1}{2x}=\dfrac{3}{2}\)

=>\(\dfrac{3}{2x}=\dfrac{3}{2}\)

=>2x=2

=>x=1(nhận)

b: ĐKXĐ: x<>0

\(\dfrac{x^2-6}{x}=x+\dfrac{3}{2}\)

=>\(x-\dfrac{6}{x}=x+\dfrac{3}{2}\)

=>\(\dfrac{-6}{x}=\dfrac{3}{2}\)

=>x=-4(nhận)

c: ĐKXĐ: x<>3/4

\(\dfrac{3x}{4x-3}=-2\)

=>-2(4x-3)=3x

=>-8x+6=3x

=>-11x=-6

=>\(x=\dfrac{6}{11}\left(nhận\right)\)

d: ĐKXĐ: x<>0

\(\dfrac{3}{8x}-\dfrac{1}{2x}=\dfrac{1}{x^2}\)

=>\(\dfrac{3}{8x}-\dfrac{4}{8x}=\dfrac{1}{x^2}\)

=>\(-\dfrac{1}{8x}=\dfrac{-8}{8x^2}\)

=>\(-\dfrac{x}{8x^2}=\dfrac{-8}{8x^2}\)

=>-x=-8

=>x=8(nhận)

e: ĐKXĐ: x<>2

\(\dfrac{x}{x-2}=\dfrac{2}{x-2}+7\)

=>\(\dfrac{x}{x-2}-\dfrac{2}{x-2}=7\)

=>-1=7(vô lý)

=>Phương trình vô nghiệm

f: ĐKXĐ: \(x\notin\left\{3;-2\right\}\)

\(\dfrac{2}{x-3}=\dfrac{1}{x+2}\)

=>2(x+2)=x-3

=>2x+4=x-3

=>2x-x=-3-4

=>x=-7(nhận)

g: ĐKXĐ: \(x\notin\left\{\dfrac{3}{2};-7\right\}\)

\(\dfrac{3x-2}{x+7}=\dfrac{6x+1}{2x-3}\)

=>\(\left(6x+1\right)\left(x+7\right)=\left(3x-2\right)\left(2x-3\right)\)

=>\(6x^2+42x+x+7=6x^2-9x-4x+6\)

=>43x+7=-13x+6

=>56x=-1

=>\(x=-\dfrac{1}{56}\left(nhận\right)\)

h: ĐKXĐ: \(x\notin\left\{0;-1\right\}\)

\(\dfrac{2x+1}{x+1}+\dfrac{2}{x}=\dfrac{2}{x\left(x+1\right)}\)

=>\(\dfrac{x\left(2x+1\right)+2\left(x+1\right)}{x\left(x+1\right)}=\dfrac{2}{x\left(x+1\right)}\)

=>\(x\left(2x+1\right)+2\left(x+1\right)=2\)

=>\(2x^2+x+2x+2=2\)

=>\(2x^2+3x=0\)

=>x(2x+3)=0

=>\(\left[{}\begin{matrix}x=0\left(loại\right)\\x=-\dfrac{3}{2}\left(nhận\right)\end{matrix}\right.\)