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13 tháng 11 2021

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

\(x^2-2x+1=9\\ \Leftrightarrow\left(x-1\right)^2=9\\ \Leftrightarrow\left[{}\begin{matrix}x-1=-3\\x-1=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x=4\end{matrix}\right.\)

\(7x^2=2x\\ \Leftrightarrow7x^2-2x=0\\ \Leftrightarrow x\left(7x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\7x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{7}\end{matrix}\right.\)

\(x^2-6x=8\\ \Leftrightarrow x^2-6x-8=0\\ \left(x^2-6x+9\right)-17=0\\ \Leftrightarrow\left(x-3\right)^2-\sqrt{17^2}=0\\ \Leftrightarrow\left(x-3-\sqrt{17}\right)\left(x-3+\sqrt{17}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-3-\sqrt{17}=0\\x-3+\sqrt{17}=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=3+\sqrt{17}\\x=3-\sqrt{17}\end{matrix}\right.\)

13 tháng 11 2021

Ơ sao làm mỗi 1 câu vậy bạn ?

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

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

c: \(=\dfrac{3x-1}{2\left(3x+1\right)}+\dfrac{3x+1}{2\left(3x-1\right)}-\dfrac{6x}{\left(3x-1\right)\left(3x+1\right)}\)

\(=\dfrac{9x^2-6x+1+9x^2+6x+1-12x}{2\left(3x-1\right)\left(3x+1\right)}=\dfrac{18x^2-12x+2}{2\left(3x-1\right)\left(3x+1\right)}\)

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

a: Ta có: \(\left(x-\dfrac{2}{5}\right)\left(x+\dfrac{2}{7}\right)>0\)

\(\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{2}{5}\\x< -\dfrac{2}{7}\end{matrix}\right.\)

5 tháng 8 2020

Bài 9 : Tìm x, biết :

a, (x - 2)(x - 3) + (x - 2) - 1 = 0

\(\Leftrightarrow\left(x-2\right)\left(x-3+1\right)-1=0\)

\(\Leftrightarrow\left(x-2\right)^2-1=0\)

\(\Leftrightarrow\left(x-2+1\right)\left(x-2-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)

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

Vậy x ={1; 3}

b, (x + 2)2 - 2x(2x + 3) = (x + 1)2

\(\Leftrightarrow\left(x+2\right)^2-\left(x+1\right)^2-2x\left(2x+3\right)=0\)

\(\Leftrightarrow\left(x+2+x+1\right)\left(x+2-x-1\right)-2x\left(2x+3\right)=0\)

\(\Leftrightarrow2x+3-2x\left(2x+3\right)=0\)

\(\Leftrightarrow\left(2x+3\right)\left(1-2x\right)=0\)

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

Vậy \(x=\left\{-\frac{3}{2};\frac{1}{2}\right\}\)
c, 6x3 + x2 = 2x

\(\Leftrightarrow6x^3+x^2-2x=0\)

\(\Leftrightarrow x\left(6x^2+x-2\right)=0\)

\(\Leftrightarrow x\left(6x^2+4x-3x-2\right)=0\)

\(\Leftrightarrow x\left[2x\left(3x+2\right)-\left(3x+2\right)\right]=0\)

\(\Leftrightarrow x\left(3x+2\right)\left(2x-1\right)=0\)

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

Vậy \(x=\left\{0;-\frac{2}{3};\frac{1}{2}\right\}\)

2 tháng 10 2021

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

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

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

\(=\left(x^2+3\right)^2\)

28 tháng 8 2020

Ít thôi -..-

a) ( 3x + 2 )( 2x + 9 )  - ( x + 3 )( 6x + 1 ) = ( x + 1 )2 - ( x + 2 )( x - 2 )

<=> 6x2 + 31x + 18 - ( 6x2 + 19x + 3 ) = x2 + 2x + 1 - ( x2 - 4 )

<=> 6x2 + 31x + 18 - 6x2 - 19x - 3 = x2 + 2x + 1 - x2 + 4

<=> 12x + 15 = 2x + 5

<=> 12x - 2x = 5 - 15

<=> 10x = -10

<=> x = -1

b) ( 2x + 3 )( x - 4 ) + ( x - 5 )( x - 2 ) = ( 3x - 5 )( x - 4 )

<=> 2x2 - 5x - 12 + x2 - 7x + 10 = 3x2 - 17x + 20

<=> 3x2 - 12x - 2 = 3x2 - 17x + 20

<=> 3x2 - 12x - 3x2 + 17x = 20 + 2

<=> 5x = 22

<=> x = 22/5

c) ( x + 2 )3 - ( x - 2 )3 - 12x( x - 1 ) = -8

<=> x3 + 6x2 + 12x + 8 - ( x3 - 6x2 + 12x - 8 ) - 12x2 + 12x = -8

<=>  x3 + 6x2 + 12x + 8 - x3 + 6x2 - 12x + 8 - 12x2 + 12x = -8

<=> 12x + 16 = -8

<=> 12x = -24

<=> x = -2

d) ( 3x - 1 )2 - 5( x + 1 ) + 6x - 3.2x + 1 - ( x - 1 )2 = 16

<=> 9x2 - 6x + 1 - 5x - 5 + 6x - 6x + 1 - ( x2 - 2x + 1 ) = 16

<=> 9x2 - 11x - 3 - x2 + 2x - 1 = 16

<=> 8x2 - 9x - 4 = 16

<=> 8x2 - 9x - 4 - 16 = 0

<=> 8x2 - 9x - 20 = 0

( Đến đây bạn có hai sự lựa chọn : 1 là vô nghiệm

                                                         2 là nghiệm vô tỉ =) )

28 tháng 8 2020

a) (3x + 2)(2x + 9) - (x + 3)(6x + 1) = (x + 1)2 - (x + 2)(x - 2)

=> 3x(2x + 9) + 2(2x + 9) - x(6x + 1) - 3(6x + 1) = x2 + 2x + 1 - x(x - 2) - 2(x - 2)

=> 6x2 + 27x + 4x + 18 - 6x2 - x - 18x - 3 = x2 + 2x + 1 - x2 + 2x - 2x + 4

=> (6x2 - 6x2) + (27x + 4x - x - 18x) + (18 - 3) = (x2 - x2) + (2x + 2x - 2x) + (1 + 4)

=> 12x + 15 = 2x + 5

=> 12x + 15  - 2x - 5 = 0

=> 10x + 10 = 0

=> 10x = -10 => x = -1

b) (2x + 3)(x - 4) + (x - 5)(x - 2) = (3x - 5)(x - 4)

=> 2x(x - 4) + 3(x - 4) + x(x - 2) - 5(x - 2) = 3x(x - 4) - 5(x - 4)

=> 2x2 - 8x + 3x - 12 + x2 - 2x - 5x + 10 = 3x2 - 12x - 5x + 20

=> (2x2 + x2) + (-8x + 3x - 2x - 5x) + (-12 + 10) = 3x2 - 17x + 20

=> 3x2 - 12x - 2 = 3x2 - 17x + 20

=> 3x2 - 12x - 2 - 3x2 + 17x - 20 = 0

=> (3x2 - 3x2) + (-12x + 17x) + (-2 - 20) = 0

=> 5x - 22 = 0

=> 5x = 22 => x = 22/5

c) (x + 2)3 - (x - 2)3 - 12x(x - 1) = -8

=> x3 + 6x2 + 12x + 8 - (x3  - 6x2 + 12x - 8) - 12x2 + 12x = -8

=> x3 + 6x2 + 12x + 8 -x3 + 6x2 - 12x + 8 - 12x2 + 12x = -8

=> (x3 - x3) + (6x2 + 6x2 - 12x2) + (12x - 12x + 12x) + (8 + 8) = -8

=> 12x + 16 = -8

=> 12x = -24

=> x = -2

Còn bài cuối làm nốt

AH
Akai Haruma
Giáo viên
4 tháng 8 2021

Bài 1 không có cơ sở để tính biểu thức.

AH
Akai Haruma
Giáo viên
4 tháng 8 2021

Bài 2:

a. 

$(6x+1)^2+(6x-1)^2-2(6x+1)(6x-1)$

$=[(6x+1)-(6x-1)]^2=2^2=4$

b.

$3(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)$

$=(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)$

$=(2^4-1)(2^4+1)(2^8+1)(2^{16}+1)$

$=(2^8-1)(2^8+1)(2^{16}+1)$
$=(2^{16}-1)(2^{16}+1)=2^{32}-1$

c.

$2C=(5^2-1)(5^2+1)(5^4+1)(5^8+1)(5^{16}+1)$

$=(5^4-1)(5^4+1)(5^8+1)(5^{16}+1)$

$=(5^8-1)(5^8+1)(5^{16}+1)$
$=(5^{16}-1)(5^{16}+1)=5^{32}-1$

$\Rightarrow C=\frac{5^{32}-1}{2}$

27 tháng 5 2017

bạn chỉ cần cố gắng là làm được