dmm

a: \(\left(x+1\right)^3+\left(x-2\right)^3=2x^3+2\left(2x-1\right)^2-9\)

\(\Leftrightarrow x^3+3x^2+3x+1+x^3-6x^2+12x-8=2x^3+2\left(4x^2-4x+1\right)-9\)

\(\Leftrightarrow2x^3-3x^2+15x-7=2x^3+8x^2-8x-7\)

\(\Leftrightarrow-11x^2+23x=0\)

\(\Leftrightarrow x\left(-11x+23\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{23}{11}\end{matrix}\right.\)

lop 8 con lic le 3

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

\(\Leftrightarrow2x^2-4x+x-2-2x^2=0\)

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

\(\Leftrightarrow-3x-2=0\)

\(\Leftrightarrow-3x=2\)

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

b) \(\left(x+3\right)\left(2x-1\right)+x^2=9\)

\(\Leftrightarrow\left(x+3\right)\left(2x-1\right)+x^2-9=0\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\3x=4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{4}{3}\end{matrix}\right.\)

`#3107.101107`

a)

`(2x + 1)(x - 2) - 2x^2 = 0`

`<=> 2x^2 - 3x - 2 - 2x^2 = 0`

`<=> -3x - 2 = 0`

`<=> -3x = 2`

`<=> x = -2/3`

Vậy, `x=-2/3`

b)

`(x + 3)(2x - 1) + x^2 = 9`

`<=> 2x^2 - 5x - 3 + x^2 = 9`

`<=> 3x^2 - 5x - 3 = 9`

`<=> 3x^2 - 3x - 12 = 0`

`<=> 3x^2 + 4x - 9x - 12 = 0`

`<=> (3x^2 - 9x) + (4x - 12) = 0`

`<=> 3x(x - 3) + 4(x - 3) = 0`

`<=> (3x + 4)(x - 3) = 0`

`<=>` TH1: `3x + 4 = 0`

`<=> 3x = -4`

`<=> x = -4/3`

TH2: `x - 3 = 0`

`<=> x = 3`

Vậy,` x \in {-4/3; 3}.`

\(\left(2x+1\right)+2\left(2x+1\right)\left(x+1\right)+\left(x-1\right)^2=9\)

\(\Leftrightarrow\left(3x\right)^2=9\)

\(\Leftrightarrow x=1;x=-1\)

`@` `\text {Ans}`

`\downarrow`

`a)`

`3x(4x-1) - 2x(6x-3) = 30`

`=> 12x^2 - 3x - 12x^2 + 6x = 30`

`=> 3x = 30`

`=> x = 30 \div 3`

`=> x=10`

Vậy, `x=10`

`b)`

`2x(3-2x) + 2x(2x-1) = 15`

`=> 6x- 4x^2 + 4x^2 - 2x = 15`

`=> 4x = 15`

`=> x = 15/4`

Vậy, `x=15/4`

`c)`

`(5x-2)(4x-1) + (10x+3)(2x-1) = 1`

`=> 5x(4x-1) - 2(4x-1) + 10x(2x-1) + 3(2x-1)=1`

`=> 20x^2-5x - 8x + 2 + 20x^2 - 10x +6x - 3 =1`

`=> 40x^2 -17x - 1 = 1`

`d)`

`(x+2)(x+2)-(x-3)(x+1)=9`

`=> x^2 + 2x + 2x + 4 - x^2 - x + 3x + 3=9`

`=> 6x + 7 =9`

`=> 6x = 2`

`=> x=2/6 =1/3`

Vậy, `x=1/3`

`e)`

`(4x+1)(6x-3) = 7 + (3x-2)(8x+9)`

`=> 24x^2 - 12x + 6x - 3 = 7 + (3x-2)(8x+9)`

`=> 24x^2 - 12x + 6x - 3 = 7 + 24x^2 +11x - 18`

`=> 24x^2 - 6x - 3 = 24x^2 + 18x -11`

`=> 24x^2 - 6x - 3 - 24x^2 + 18x + 11 = 0`

`=> 12x +8 = 0`

`=> 12x = -8`

`=> x= -8/12 = -2/3`

Vậy, `x=-2/3`

`g)`

`(10x+2)(4x- 1)- (8x -3)(5x+2) =14`

`=> 40x^2 - 10x + 8x - 2 - 40x^2 - 16x + 15x + 6 = 14`

`=> -3x + 4 =14`

`=> -3x = 10`

`=> x= - 10/3`

Vậy, `x=-10/3`

Hello các bạn còn đó ko?

Lời giải:

$(2x+1)(2x-1)-(2x-3)^2=-9$

$\Leftrightarrow (4x^2-1)-(4x^2-12x+9)=-9$

$\Leftrightarrow 4x^2-1-4x^2+12x-9=-9$

$\Leftrightarrow 12x-10=-9$
$\Leftrightarrow 12x=1$

$\Leftrightarrow x=\frac{1}{12}$

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

\(\Leftrightarrow\left(4x^2-1\right)-\left(4x^2-12x+9\right)^2=-9\)

\(\Leftrightarrow4x^2-1-4x^2+12x-9=-9\)

\(\Leftrightarrow12x-10=-9\)

\(\Leftrightarrow12x=1\)

\(\Leftrightarrow x=\dfrac{1}{12}\)

Vậy: \(x=\dfrac{1}{12}\)

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

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

b) \(x^2-1=0\Rightarrow\left(x-1\right)\left(x+1\right)=0\)

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

c) \(x^2-9=0\Rightarrow\left(x-3\right)\left(x+3\right)=0\)

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

d) \(\Rightarrow\left(2x-4\right)\left(2x+4\right)=0\)

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

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

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\\x=-\dfrac{3}{5}\end{matrix}\right.\)

\(a,\Leftrightarrow\left(x-2\right)\left(5x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{2}{5}\end{matrix}\right.\\ b,\Leftrightarrow2x^2+2x-x^2+4x-4-6=0\\ \Leftrightarrow x^2+6x-10=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-3+\sqrt{19}\\x=-3-\sqrt{19}\end{matrix}\right.\\ c,\Leftrightarrow2x^2-2x+9x-9=0\\ \Leftrightarrow\left(2x+9\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{9}{2}\\x=1\end{matrix}\right.\)

`@` `\text {Ans}`

`\downarrow`

`a,`

`x^2 + 2x + 1 = 9`

`=> x^2 + 2x + 1 - 9 = 0`

`=> x^2 + 2x - 8 = 0`

`=> x^2 + 4x - 2x - 8 = 0`

`=> (x^2 + 4x) - (2x + 8) = 0`

`=> x(x + 4) - 2(x + 4) = 0`

`=> (x-2)(x+4) = 0`

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

`=>`\(\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)

Vậy, `x \in {2; 4}`

`b,`

`x^2 - 1 = 15`

`=> x^2 = 15 + 1`

`=> x^2 = 16`

`=> x^2 = (+-4)^2`

`=> x = +-4`

Vậy, `x \in {4; -4}`

`c)`

`19 - 2x^2 = 1`

`=> 2x^2 = 19 - 1`

`=> 2x^2 = 18`

`=> x^2 = 18 \div 2`

`=> x^2 = 9`

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

`=> x = +-3`

Vậy, `x \in {3; -3}.`

1. a. 9.(x-2)+1=2-2x

=> 9x-18+1=2-2x

=> 9x+2x = 2+18-1

=> 11x = 19

=> x=19/11

b. 9|x| = 45

=> |x| = 5

=> x=5 hoặc x=-5

c. 9:|x|=3

=> |x| = 3

=> x=3 hoặc x=-3

d. 5.(1-x)+2x=9-2x

=> 5-5x+2x=9-2x

=> -5x+2x+2x=9-5

=> -x=4

=> x=-4

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