kh hiểu bn ơi

`4x=2+xx+1x<=>4x=2+3x<=>4x-3x=2<=>1x=2<=>x=2`

bạn tách từng bài ra bn

tl câu hỏi trên cho mik ik

a: (x^2+9)(9x^2-1)=0

=>9x^2-1=0

=>x^2=1/9

=>x=1/3 hoặc x=-1/3

b: (4x^2-9)(2^(x-1)-1)=0

=>4x^2-9=0 hoặc 2^(x-1)-1=0

=>x^2=9/4 hoặc x-1=0

=>x=1;x=3/2;x=-3/2

c: (3x+2)(9-x^2)=0

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

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

d: (3x+3)^2(4x-4^2)=0

=>3x+3=0 hoặc 4x-16=0

=>x=4 hoặc x=-1

e: \(2^{\left(x-5\right)\left(x+2\right)}=1\)

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

=>x-5=0 hoặc x+2=0

=>x=5 hoặc x=-2

( 2 x   –   3 ) 2   –   4 x 2   +   9   =   0     ⇔   ( 2 x   –   3 ) 2   –   ( 4 x 2   –   9 )   =   0     ⇔   ( 2 x   –   3 ) 2   –   ( ( 2 x ) 2   –   3 2 )   =   0     ⇔   ( 2 x   –   3 ) 2   –   ( 2 x   –   3 ) ( 2 x   +   3 )   =   0

ó (2x – 3)(2x – 3 – 2x – 3) = 0

ó (2x – 3)(-6) = 0

ó 2x – 3 = 0

ó x = 3 2

Đáp án cần chọn là: C

a)4x²-9=0

⇔ (2x-3)(2x+3)=0

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

b)(x+5)²-(x-1)²=0

⇔ (x+5-x+1)(x+5+x-1)=0

⇔ 12(x+2)=0

⇔ x=-2

c)x²-6x-7=0

⇔ x²-7x+x-7=0

⇔ x(x-7)+(x-7)=0

⇔ (x-7)(x+1)=0

\(\Leftrightarrow\left\{{}\begin{matrix}x=7\\x=-1\end{matrix}\right.\)

d)(x+1)²-(2x-1)²=0

⇔ (x+1-2x+1)(x+1+2x-1)=0

⇔3x(2-x)=0

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

a, 4x² - 9 = 0

<=> 4x² = 9

<=> x² = \(\dfrac{9}{4}\) => x = \(\sqrt{\dfrac{9}{4}}\)

b, (x + 5 )² - ( x - 1 )² = 0

<=> ( x+5-x+1 )(x+5+x-1) = 0

<=> 6(2x+4) = 0

<=> 12x+24=0

<=> 12x = -24

<=> x = -2

c, x²-6x-7=0

<=> x²+x-7x-7=0

<=> x(x+1)-7(x+1)=0

<=> (x-7)(x+1)=0

=> x+7=0 hoặc x+1=0

+ x-7=0 => x=7

+ x+1=0 => x=-1

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

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

<=> (-x+2).3x=0

=> x=0 hoặc (-x+2).3=0

+ (-x+2).3=0 => -3x+6=0 => x=-2

b) (x +5)² -(x -1)²=0

<=> [(x +5) -(x -1)][(x +5) +(x -1)]=0

<=> (x +5 -x +1)(x +5 +x -1)=0

<=> 6(2x+4)=0 <=>12(x +2)=0

=> x +2=0=> x=-2

vậy x= -2

c) x² -6x -7=0

<=> x² -7x +x -7=0

<=> (x² +x)( -7x -7)=0

<=> x(x +1).-7(x +1)=0

<=> (x +1)(x -7)=0

<=> \(\left\{{}\begin{matrix}x+1=0\\x-7=0\end{matrix}\right.< =>\left\{{}\begin{matrix}x=-1\\x=7\end{matrix}\right.\)

Vậy S={-1; 7}

d) (x +1)² -(2x -1)²=0

<=> [(x -1)-(2x -1)][(x -1)+(2x -1)]=0

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

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

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

Vậy S={0; \(\dfrac{2}{3}\)}

\(a,\Leftrightarrow\left(2x-3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-2\end{matrix}\right.\\ b,\Leftrightarrow x^3-27-x^3+4x=1\\ \Leftrightarrow4x=28\Leftrightarrow x=7\\ c,\Leftrightarrow4x^2-4x-8=0\\ \Leftrightarrow x^2-x-2=0\\ \Leftrightarrow\left(x-2\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\\ d,\Leftrightarrow2x^2+6x+x+3=0\\ \Leftrightarrow\left(x+3\right)\left(2x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{1}{2}\end{matrix}\right.\)

\(a,\Leftrightarrow\left(x-2\right)\left(3x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-2\right)^3=0\Leftrightarrow x-2=0\Leftrightarrow x=2\\ c,\Leftrightarrow\left(4x-3x-3\right)\left(4x+3x+3\right)=0\\ \Leftrightarrow\left(x-3\right)\left(7x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{7}\end{matrix}\right.\\ d,\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

\(1,\Leftrightarrow x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=9\\x=0\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\Leftrightarrow-4x=7\Leftrightarrow x=-\dfrac{7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\Leftrightarrow5x=15\Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(x-7\right)\left(3x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-\dfrac{4}{3}\end{matrix}\right.\)

\(7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\\ 8,\Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=4\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\\ 11,\Leftrightarrow\left(4x-3\right)\left(3-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{2}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\Leftrightarrow-10x=3\Leftrightarrow x=-\dfrac{3}{10}\)

\(1,\Leftrightarrow x\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\\ \Leftrightarrow-4x=7\\ \Leftrightarrow x=\dfrac{-7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\\ \Leftrightarrow5x=15\\ \Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\)

\(5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(3x+4\right)\left(x-7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=7\end{matrix}\right.\\ 7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

\(8,\Leftrightarrow10x\left(x-4\right)+2\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\\ \Leftrightarrow-5x=0\\ \Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

\(11,\Leftrightarrow\left(2x-3\right)\left(4x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{4}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\\ \Leftrightarrow-10x=3\\ \Leftrightarrow x=-\dfrac{3}{10}\)

1) \(x^2-9x=0\Rightarrow x\left(x-9\right)=0\Rightarrow x=0;9\)

2) \(x\left(x-4\right)-x^2=7\Rightarrow-4x=7\Rightarrow x=-\dfrac{7}{4}\)

3) \(3x+2\left(x-5\right)=5\Rightarrow5x-10=5\Rightarrow5x=15\Rightarrow x=3\)

4) \(25x^2-1=0\Rightarrow x^2=\dfrac{1}{25}\Rightarrow x=\pm\dfrac{1}{5}\)

5) \(3x\left(x-2\right)-5\left(x-2\right)=0\Rightarrow\left(x-2\right)\left(3x-5\right)=0\Rightarrow x=2;\dfrac{5}{3}\)

6) \(3x\left(x-7\right)+4\left(x-7\right)\Rightarrow\left(3x+4\right)\left(x-7\right)=0\Rightarrow x=-\dfrac{4}{3};7\)

7) \(4x^2-9=0\Rightarrow x^2=\dfrac{9}{4}\Rightarrow x=\pm\dfrac{3}{2}\)

8) \(10x\left(x-4\right)+2x-8=0\Rightarrow2\left(x-4\right)\left(5x+1\right)=0\Rightarrow x=4;-\dfrac{1}{5}\)

9) \(x\left(2x-5\right)-2x^2=0\Rightarrow x\left(2x-5-2x=0\right)\Rightarrow x=0\)

10) \(2x^2-4x=0\Rightarrow2x\left(x-2\right)=0\Rightarrow x=0;2\)

11) \(2x\left(3-4x\right)+3\left(4x-3\right)=0\Rightarrow2x\left(4x-3\right)-3\left(4x-3\right)=0\Rightarrow\left(4x-3\right)\left(2x-3\right)=0\Rightarrow x=\dfrac{3}{4};\dfrac{3}{2}\)

12) \(2x\left(x-5\right)-2x^2=3\Rightarrow-10x=3\Rightarrow x=-\dfrac{3}{10}\)

\(\left(4x^2\right)^8\cdot\left(4x^2\right)\cdot\left(4x^2\right)^2\)

\(=\left[\left(2x\right)^2\right]^8\cdot\left[\left(2x\right)^2\right]^4\cdot\left[\left(2x\right)^2\right]^2\)

\(=\left(2x\right)^{16}\cdot\left(2x\right)^8\cdot2x^4\)

\(=\left(2x\right)^{28}\)