a,(x2 - 2x + 1)(x - 1)
CD
Những câu hỏi liên quan
Bài 1: Tính:
a) x2(x-2x3); b) (x2+1)(5-x); c) (x-2)(x2+3x-4); d) (x-2)(x-x2+4); e) (x2-1)(x2+2x); f) (2x-1)(3x+2)(3-x)
Bài 2: Tính:
a) (x-2y)2; b) (2x2+3)3; c) (x-2)(x2+2x+4); d) (2x-1)3
Bài 3: Rút gọn biểu thức:
a) (6x+1)2+(6x-1)2-2(1+6x)(6x-1); b) 3(22+1)(24+1)(28+1)(216+1); c) x(2x2-3)-x2(5x+1)+x2; d) 3x(x-2)-5x(1-x)-8(x2-3)
Bài 4: Tính nhanh:
a) 1012; b) 97.103; c) 772+232+77.46; d) 1052-52; e) A (x-y)(x2+xy+y2)+2y3 tại x dfrac{2}{3} và y dfrac{1}{3}
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Bài 1: Tính:
a) x2(x-2x3); b) (x2+1)(5-x); c) (x-2)(x2+3x-4); d) (x-2)(x-x2+4); e) (x2-1)(x2+2x); f) (2x-1)(3x+2)(3-x)
Bài 2: Tính:
a) (x-2y)2; b) (2x2+3)3; c) (x-2)(x2+2x+4); d) (2x-1)3
Bài 3: Rút gọn biểu thức:
a) (6x+1)2+(6x-1)2-2(1+6x)(6x-1); b) 3(22+1)(24+1)(28+1)(216+1); c) x(2x2-3)-x2(5x+1)+x2; d) 3x(x-2)-5x(1-x)-8(x2-3)
Bài 4: Tính nhanh:
a) 1012; b) 97.103; c) 772+232+77.46; d) 1052-52; e) A= (x-y)(x2+xy+y2)+2y3 tại x= \(\dfrac{2}{3}\) và y= \(\dfrac{1}{3}\)
Bạn chú ý đăng lẻ câu hỏi! 1/
a/ \(=x^3-2x^5\)
b/\(=5x^2+5-x^3-x\)
c/ \(=x^3+3x^2-4x-2x^2-6x+8=x^3=x^2-10x+8\)
d/ \(=x^2-x^3+4x-2x+2x^2-8=3x^2-x^3+2x-8\)
e/ \(=x^4-x^2+2x^3-2x\)
f/ \(=\left(6x^2+x-2\right)\left(3-x\right)=17x^2+5x-6-6x^3\)
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Giải pt : a) 2/-x2+6x-8 - x-1/x-2 = x+3/x-4
b) 2/x3-x2-x+1 = 3/1-x2 - 1/x+1
c) x+2/x-2 - 2/x2-2x = 1/x
d) 5/-x2+5x-6 + x+3/2-x = 0
e) x/2x+2 - 2x/x2-2x-3 = x/6-2x
f) 1/x-1 - 3x2/x3-1 = 2x/x2+x-1
a) (x2 - 5x)2 + 10(x2 - 5x) + 24 = 0
b) (2x + 1)2 - 2x - 1 = 2
c) x(x - 1)(x2 - x + 1) - 6 = 0
d) (x2 + 1)2 + 3x(x2 + 1) + 2x2 = 0
a) Ta có: \(\left(x^2-5x\right)^2+10\left(x^2-5x\right)+24=0\)
\(\Leftrightarrow\left(x^2-5x\right)^2+4\left(x^2-5x\right)+6\left(x^2-5x\right)+24=0\)
\(\Leftrightarrow\left(x^2-5x\right)\left(x^2-5x+4\right)+6\left(x^2-5x+4\right)=0\)
\(\Leftrightarrow\left(x^2-5x+6\right)\left(x^2-5x+4\right)=0\)
\(\Leftrightarrow\left(x^2-2x-3x+6\right)\left(x^2-x-4x+4\right)=0\)
\(\Leftrightarrow\left[x\left(x-2\right)-3\left(x-2\right)\right]\left[x\left(x-1\right)-4\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\\x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=3\\x=4\end{matrix}\right.\)
Vậy: S={1;2;3;4}
b) Ta có: \(\left(2x+1\right)^2-2x-1=2\)
\(\Leftrightarrow\left(2x+1\right)^2-\left(2x+1\right)-2=0\)
\(\Leftrightarrow\left(2x+1\right)^2-2\left(2x+1\right)+\left(2x+1\right)-2=0\)
\(\Leftrightarrow\left(2x+1\right)\left(2x+1-2\right)+\left(2x+1-2\right)=0\)
\(\Leftrightarrow\left(2x+1+1\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left(2x+2\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+2=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-2\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{-1;\dfrac{1}{2}\right\}\)
c) Ta có: \(x\left(x-1\right)\left(x^2-x+1\right)-6=0\)
\(\Leftrightarrow x\left(x^3-x^2+x-x^2+x-1\right)-6=0\)
\(\Leftrightarrow x\left(x^3-2x^2+2x-1\right)-6=0\)
\(\Leftrightarrow x^4-2x^3+2x^2-x-6=0\)
\(\Leftrightarrow x^4-2x^3+2x^2-4x+3x-6=0\)
\(\Leftrightarrow x^3\left(x-2\right)+2x\left(x-2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3+2x+3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3-x+3x+3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[x\left(x^2-1\right)+3\left(x+1\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left[x\left(x-1\right)\left(x+1\right)+3\left(x+1\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+1\right)\left(x^2-x+3\right)=0\)
mà \(x^2-x+3>0\forall x\)
nên (x-2)(x+1)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
Vậy: S={2;-1}
d) Ta có: \(\left(x^2+1\right)^2+3x\left(x^2+1\right)+2x^2=0\)
\(\Leftrightarrow\left(x^2+1\right)^2+2x\left(x^2+1\right)+x\left(x^2+1\right)+2x^2=0\)
\(\Leftrightarrow\left(x^2+1\right)\left(x^2+1+2x\right)+x\left(x^2+1+2x\right)=0\)
\(\Leftrightarrow\left(x+1\right)^2\cdot\left(x^2+x+1\right)=0\)
mà \(x^2+x+1>0\forall x\)
nên x+1=0
hay x=-1
Vậy: S={-1}
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1 Trong các phương trình sau, phương trình nào vô nghiệm:
A. x2 – 2x + 2 = 0 B. x2 – 2x + 1 = 0
C. x2 – 2x = 0 D. 2x – 10 = 2x – 10
2 Phương trình nào sau đây có 1 nghiệm :
A. x2 – 3 x = 0 B. 2x + 1 =1 +2x
C. x ( x – 1 ) = 0 D. (x + 2)(x2 + 1) = 0
a. x2(x – 2x3) b. (x2 + 1)(5 – x) c. (x – 2)(x2 + 3x – 4) d. (x – 2)(x – x2 + 4) e. (x2 – 1)(x2 + 2x) f. (2x – 1)(3x + 2)(3 – x) g. (x + 3)(x2 + 3x – 5) h. (xy – 2).(x3 – 2x – i. (5x3 – x2 + 2x – 3).(4x2 – x + 2
a: \(=x^3-2x^5\)
e: \(=x^4+2x^3-x^2-2x\)
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a)A=3x(2/3x2-3x4)+(3x2)(x3-1)+(-2+9).x2-12
b)B=x(2x3+x+2)-2x2(x2+1)+x2-2x+1
c)C=x.(2x+1)-x2(x+2)+x3-x+3
a, \(A=2x^3-9x^5+3x^5-3x^2+7x^2-12=-6x^5+2x^3+4x^2-12\)
b, \(B=2x^4+x^2+2x-2x^3-2x^2+x^2-2x+1=2x^4-2x^3+1\)
c, \(C=2x^2+x-x^3-2x^2+x^3-x+3=3\)
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a. x+1/x-2 - x/x+2 + 8/x2 -4
b. x-3/x+1 - x+2/x-1 + 8x/x2 -1
c. x+2/x2-2x + 2/x2+2x + 3x+2/x2-4
d. 4/x - 12/x2+3x + 5/x+3
a: \(=\dfrac{x^2+3x+2-x^2+2x+8}{\left(x-2\right)\left(x+2\right)}=\dfrac{5x+10}{\left(x-2\right)\left(x+2\right)}=\dfrac{5}{x-2}\)
b: \(=\dfrac{x^2-4x+3-x^2-3x-2+8x}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}=\dfrac{1}{x-1}\)
c: \(=\dfrac{x+2}{x\left(x-2\right)}+\dfrac{2}{x\left(x+2\right)}+\dfrac{3x+2}{\left(x+2\right)\left(x-2\right)}\)
\(=\dfrac{x^2+2x+2x-4+3x+2}{x\left(x-2\right)\left(x+2\right)}=\dfrac{x^2+7x-2}{x\left(x-2\right)\left(x+2\right)}\)
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a,
\(\dfrac{x+1}{x-2}-\dfrac{x}{x+2}+\dfrac{8}{x^2-4}\\ =\dfrac{x^2+3x+2-x^2+2x+8}{\left(x-2\right)\left(x+2\right)}=\dfrac{5x+10}{\left(x-2\right)\left(x+2\right)}=\dfrac{5\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{5}{x-2}\)
b,
\(\dfrac{x-3}{x+1}-\dfrac{x+2}{x-1}+\dfrac{8x}{x^2-1}\\ =\dfrac{x^2-4x+3-x^2-3x-2+8x}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{1}{x-1}\)
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rút gọn
a) x2 (5x3-x-6)
b)(6x2-5) (2x+3)
c)(x2-2x+1):(x-1)
d)3x (x2+x-1)
PK
28 tháng 7 2021 lúc 9:29
ai giúp mình giải bài này với được k mình đang cần gấp ( xin cảm ơn)Bài 1: a,√3x+4−√2x+1√x+3b, √2x−5+√x+2√2x+1c, √x+4−√1−x√1−2xd, √x+95−√2x+4Bài 2:a,√x+4√x+45x+2b, √x2−2x+1+√x2+4x+44c, √x+2√x−1+√x−2√x−12d,√x−2+√2x−5+√x+2+3√2x−57√2Bài 3:a, x2−7x6√x+5−30
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ai giúp mình giải bài này với được k mình đang cần gấp ( xin cảm ơn)
Bài 1:
a,√3x+4−√2x+1=√x+3
b, √2x−5+√x+2=√2x+1
c, √x+4−√1−x=√1−2x
d, √x+9=5−√2x+4
Bài 2:
a,√x+4√x+4=5x+2
b, √x2−2x+1+√x2+4x+4=4
c, √x+2√x−1+√x−2√x−1=2
d,√x−2+√2x−5+√x+2+3√2x−5=7√2
Bài 3:
a, x2−7x=6√x+5−30
Thuc hien phep nhan vs da thuc:
a, (x2 + 2x+1)(x+1)
b, (x3-x2+2x-1)(5-x)
c, (x-5)(x3-x2+2x-1)
Giải:
a) \(\left(x^2+2x+1\right)\left(x+1\right)\)
\(=x^2.x+2x.x+1.x+x^2.1+2x.1+1.1\)
\(=x^3+2x^2+x+x^2+2x+1\)
\(=x^3+3x^2+3x+1\)
b) \(\left(x^3-x^2+2x-1\right)\left(5-x\right)\)
\(=x^3.5-x^2.5+2x.5-1.5+x^3.\left(-x\right)-x^2.\left(-x\right)+2x.\left(-x\right)-1.\left(-x\right)\)
\(=5x^3-5x^2+10x-5-x^4+x^3-2x^2+x\)
\(=6x^3-7x^2+11x-5-x^4\)
c) \(\left(x-5\right)\left(x^3-x^2+2x-1\right)\)
\(=x.x^3-5.x^3+x.\left(-x^2\right)-5.\left(-x^2\right)+x.2x-5.2x+x.\left(-1\right)-5.\left(-1\right)\)
\(=x^4-5x^3-x^3+5x^2+2x^2-10x-x+5\)
\(=x^4-6x^3+7x^2-11x+5\)
Chúc bạn học tốt!!!
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lớp 8 Phạm Hoàng Giang không chơi kiểu lớp 7
đúng làm 8 mà làm
\(A=\left(x^2+2x+1\right)\left(x+1\right)=\left(x+1\right)^2\left(x+1\right)=\left(x+1\right)^3\)
\(A=x^3+3x^2+3x+1\)
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