Ta có \(\dfrac{2x-3}{5}=\dfrac{3y+2}{7}=\dfrac{z-1}{3}=\dfrac{4x-6}{10}=\dfrac{6y+4}{14}=\dfrac{7z-7}{21}\)

Áp dụng t/c dtsbn:

\(\dfrac{4x-6}{10}=\dfrac{6y+4}{14}=\dfrac{7z-7}{21}=\dfrac{\left(4x-6y+7z\right)-6-4-7}{10-14+21}=\dfrac{68-17}{17}=3\\ \Rightarrow\left\{{}\begin{matrix}2x-3=15\\3y+2=21\\z-1=9\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=9\\y=\dfrac{19}{3}\\z=10\end{matrix}\right.\)

1) \(2x\left(x-5\right)+\left(x-2\right)\left(x+3\right)=2x^2-10x+x^2+3x-2x-6=3x^2-9x-6\)

2) \(\left(2x-5\right)\left(1-x\right)-\left(x-3\right)\left(-2x\right)=2x-2x^2-5+5x+2x^2-6x=x-5\)

3) \(\left(4x-3\right)\left(4x-3\right)-\left(3x+2\right)\left(3x-2\right)=\left(4x-3\right)^2-9x^2+4=16x^2-24x+9-9x^2+4\)

\(=7x^2-24x+13\)

4) \(\left(2x-1\right)\left(2x+1\right)\left(2x+1\right)-4\left(x^2+1\right)=\left(2x-1\right)[\left(2x+1\right)^2]-4x^2-4\)

\(=\left(2x-1\right)\left(4x^2+4x+4\right)-4x^2-4=8x^3+8x^2+8x-4x^2-4x-4-4x^2-4=8x^3+4x-8\)

5) \(3x\left(2x-8\right)-\left(2-6x\right)\left(5+x\right)=6x^2-24x-10-2x+30x+6x^2=12x^2+4x-10\)

6) \(x\left(3x-18\right)-3\left(x-4\right)\left(x-2\right)+8=3x^2-18x-3x^2+6x+12x-24+8=-16\)

7) \(\left(x+2\right)\left(x^2-2x+4\right)-x^2\left(x-2\right)-2x^2=x^3+8-x^3+2x^2-2x^2=8\)

68-4x=2x+2¹⁹/2¹³

68-4x=2x+2⁶

68=2x+2⁶+4x

68=6x+2⁶

68=6x+64

6x=68-64

6x=4

x=4/6

x=2/3

cậu vô câu hỏi tương tự lỡ có đó

điên à Thảo Jackson

Phân tích đa thức thành nhân tử(tách hạng tử)
1)x^2+2x-3=x^2-x+3x-3=x(x-1)+3(x-1)=(x-1)(x+3)
2)x^2-5x+6=x^2-2x-3x+6=x(x-2)-3(x-2)=(x-2)(x-3)
3)x^2+7x+12=(x+3)(x+4)
4)x^2-x-12=(x-4)(x+3)
5)3x^2+3x-36=3[(x-3)(x+4)]
6)5x^2-5x-10=5[(x-2)(x+1) ]       
7)3x^2-7x-6=(x-3)(3x+2)
8)4x^2+4x-3=4x^2+6x-2x-3=(2x-1)(2x+3)
9)8x^2-2x-3=8x^2+4x-6x-3=(4x-3)(2x+1)
 

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

2: \(x^2-5x+6=\left(x-2\right)\left(x-3\right)\)

3: \(x^2+7x^2+12x=4x\left(2x+3\right)\)

4: \(x^2-x-12=\left(x-4\right)\left(x+3\right)\)

5: \(3x^2+3x-36=3\left(x^2+x-12\right)=3\left(x+4\right)\left(x-3\right)\)

6: \(5x^2-5x-10=5\left(x^2-x-2\right)=5\left(x-2\right)\left(x+1\right)\)

??????????????????

a, \(3x=5y=7z\Rightarrow\frac{x}{\frac{1}{3}}=\frac{y}{\frac{1}{5}}=\frac{z}{\frac{1}{7}}\)

\(\Rightarrow\frac{2x}{\frac{2}{3}}=\frac{y}{\frac{1}{5}}=\frac{3z}{\frac{3}{7}}\)
Áp dụng t/c

\(\Rightarrow\frac{2x}{\frac{2}{3}}=\frac{y}{\frac{1}{5}}=\frac{3z}{\frac{3}{7}}=\frac{2x-y+3z}{\frac{2}{3}-\frac{1}{5}+\frac{3}{7}}=\frac{188}{\frac{105}{94}}=210\)

\(\frac{x}{\frac{1}{3}}=210\Rightarrow x=70\)

\(\frac{y}{\frac{1}{5}}=210\Rightarrow y=42\)

\(\frac{z}{\frac{1}{7}}=210\Rightarrow z=30\)

`Answer:`

a. \(5x-[2x+1-\left(2x-3\right)-\left(4x+1\right)]\)

\(=5x-\left(2x+1-2x+3-4x-1\right)\)

\(=5x-2x-1+2x-3+4x+1\)

\(=\left(5x-2x+2x+4x\right)+\left(-3-1+1\right)\)

\(=9x-3\)

b. \(\left(-3x^2+2x-1\right)+\left(4x^2-2x+3\right)\)

\(=-3x^2+2x-1+4x^2-2x+3\)

\(=\left(-3x^2+4x^2\right)+\left(2x-2x\right)+\left(-1+3\right)\)

\(=x^2+2\)