a) 9-64x^2=0

=>  64x^2  = 8

=>  \(x^2=\frac{8}{64}=\frac{1}{8}\)

=> \(x=\frac{1}{\sqrt{8}}\)

 b )   25x^2  -  3  =  0

=>  25x^2  =  3 

=>  \(x^2=\frac{3}{25}\)    

=>  \(x=\frac{\sqrt{3}}{5}\)           

C)  7  -  16x^2  =0

=>  16x^2   =  7

=>  \(x^2=\frac{7}{16}\)       

=>   \(x=\frac{\sqrt{7}}{4}\)    

d)  4x^2  -  (x-4)^2 = 0

=>  4x^2  - x^2 + 8x - 16 =0

=>  3x^2 + 8x -16  =  0 

=> ( 3x^2 + 12x ) - ( 4x  +16 ) =  0 

=>  3x( x + 4 ) - 4( x + 4 ) =  0 

=>( x + 4 )( 3x - 4 ) =  0 

=>   \(\orbr{\begin{cases}x+4=0\\3x-4=0\end{cases}}\)    

=>  \(\orbr{\begin{cases}x=-4\\x=\frac{4}{3}\end{cases}}\)         

e)  ( 3x + 4 )^2 - ( 2x - 5 )^2 = 0

=>  ( 3x + 4 + 2x - 5 )( 3x + 4 - 2x + 5 )  = 0

=>   ( 5x -1 ) ( x + 9 )  = 0 

=>  \(\orbr{\begin{cases}5x-1=0\\x+9=0\end{cases}}\)     

=> \(\orbr{\begin{cases}x=\frac{1}{5}\\x=-9\end{cases}}\)            

Trả lời:

a, \(9-64x^2=0\)

\(\Leftrightarrow\left(3-8x\right)\left(3+8x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3-8x=0\\3+8x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{8}\\x=-\frac{3}{8}\end{cases}}}\)

Vậy x = 3/8; x = - 3/8 là nghiệm của pt.

b, \(25x^2-3=0\)

\(\Leftrightarrow\left(5x-\sqrt{3}\right)\left(5x+\sqrt{3}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}5x-\sqrt{3}=0\\5x+\sqrt{3}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{\sqrt{3}}{5}\\x=-\frac{\sqrt{3}}{5}\end{cases}}}\)

Vậy \(x=\pm\frac{\sqrt{3}}{5}\)

c, \(7-16x^2=0\)

\(\Leftrightarrow\left(\sqrt{7}-4x\right)\left(\sqrt{7}+4x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{7}-4x=0\\\sqrt{7}+4x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{\sqrt{7}}{4}\\x=-\frac{\sqrt{7}}{4}\end{cases}}}\)

Vậy \(x=\pm\frac{\sqrt{7}}{4}\)

d, \(4x^2-\left(x-4\right)^2=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x+4=0\\3x-4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-4\\x=\frac{4}{3}\end{cases}}}\)

Vậy x = - 4; x = 4/3 là nghiệm của pt.

e, \(\left(3x+4\right)^2-\left(2x-5\right)^2=0\)

\(\Leftrightarrow\left(3x+4-2x+5\right)\left(3x+4+2x-5\right)=0\)

\(\Leftrightarrow\left(x+9\right)\left(5x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+9=0\\5x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-9\\x=\frac{1}{5}\end{cases}}}\)

Vậy x = - 9; x = 1/5 là nghiệm của pt.

a) \(\sqrt{\left(2x-3\right)^2}=7\)

\(\Leftrightarrow\left|2x-3\right|=7\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}2x=10\\2x=-4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

b) \(\sqrt{64x+128}-\sqrt{25x+50}+\sqrt{4x+8}=20\left(đk:x\ge-2\right)\)

\(\Leftrightarrow8\sqrt{x+2}-5\sqrt{x+2}+2\sqrt{x+2}=20\)

\(\Leftrightarrow5\sqrt{x+2}=20\)

\(\Leftrightarrow\sqrt{x+2}=4\Leftrightarrow x+2=16\Leftrightarrow x=14\left(tm\right)\)

c) \(\sqrt{x^2-9}-3\sqrt{x-3}=0\left(đk:x\ge3\right)\)

\(\Leftrightarrow\sqrt{\left(x-3\right)\left(x+3\right)}-3\sqrt{x-3}=0\)

\(\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}-3\right)=0\)

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

a. \(\sqrt{\left(2x-3\right)^2}=7\)

<=> \(\left|2x-3\right|=7\)

<=> \(\left[{}\begin{matrix}2x-3=7\left(x\ge\dfrac{3}{2}\right)\\-2x+3=7\left(x< \dfrac{3}{2}\right)\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}2x=10\\-2x=4\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=5\left(TM\right)\\x=-2\left(TM\right)\end{matrix}\right.\)

b. \(\sqrt{64x+128}-\sqrt{25x+50}+\sqrt{4x+8}=20\)  ĐK: \(x\ge-2\)

<=> \(\sqrt{64\left(x+2\right)}-\sqrt{25\left(x+2\right)}+\sqrt{4\left(x+2\right)}-20=0\)

<=> \(8\sqrt{x+2}-5\sqrt{x+2}+2\sqrt{x+2}-20=0\)

<=> \(\sqrt{x+2}.\left(8-5+2\right)-20=0\)

<=> \(5\sqrt{x+2}=20\)

<=> \(\sqrt{x+2}=4\)

<=> \(\left(\sqrt{x+2}\right)^2=4^2\)

<=> \(\left|x+2\right|=16\)

<=> \(\left[{}\begin{matrix}x+2=16\left(x\ge-2\right)\\x+2=-16\left(x< -2\right)\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=14\left(TM\right)\\x=-18\left(TM\right)\end{matrix}\right.\)

c. \(\sqrt{x^2-9}-3\sqrt{x-3}=0\)             ĐK: \(x\ge3\)

<=> \(\sqrt{\left(x-3\right)\left(x+3\right)}-3\sqrt{x-3}=0\)

<=> \(\sqrt{x-3}.\sqrt{x+3}-3\sqrt{x-3}=0\)

<=> \(\left(\sqrt{x+3}-3\right).\sqrt{x-3}=0\)

<=> \(\left[{}\begin{matrix}\sqrt{x+3}-3=0\\\sqrt{x-3}=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=6\\x=3\end{matrix}\right.\)

a) I2x-3I=7

        2x-3=7     =>x=5

        2x-3=-7     =>x=-2

b) \(8\sqrt{3x}-5\sqrt{3x}+2\sqrt{3x}=20\)

   5\(\sqrt{3x}=20\)

   3x=16  =>x=16/3

c)  vì câu c dài nên mình chỉ cho đáp án thôi là 0,3,6

vì \(\sqrt{ }\) của 1 số luôn dương nên 3,6 thỏa mãn 

a) Ta có: \(2\sqrt{9x-27}-\dfrac{1}{5}\sqrt{25x-75}-\dfrac{1}{7}\sqrt{49x-147}=20\)

\(\Leftrightarrow6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20\)

\(\Leftrightarrow4\sqrt{x-3}=20\)

\(\Leftrightarrow x-3=25\)

hay x=28

b) Ta có: \(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\)

\(\Leftrightarrow3\sqrt{x+2}-5\sqrt{x+2}+4\sqrt{x+2}=6\)

\(\Leftrightarrow2\sqrt{x+2}=6\)

\(\Leftrightarrow x+2=9\)

hay x=7

Bài 1:

\(A=-x^2-2x+9\)

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

\(A=-\left(x^2+2x+1-10\right)\)

\(A=-\left(x+1\right)^2+10\)

Vì \(-\left(x+1\right)^2\le0\) với mọi x

\(\Rightarrow-\left(x+1\right)^2+10\le10\)

\(\Rightarrow Amax=10\Leftrightarrow x=-1\)

\(B=-9x^2+6x+25\)

\(B=-\left(9x^2-6x-25\right)\)

\(B=-\left[\left(3x\right)^2-2.3x+1-26\right]\)

\(B=-\left(3x-1\right)^2+26\)

Vì \(-\left(3x-1\right)^2\le0\) với mọi x

\(\Rightarrow-\left(3x-1\right)^2+26\le26\)

\(\Rightarrow Bmax=26\Leftrightarrow3x-1=0\Rightarrow x=\dfrac{1}{3}\)

\(C=-x^2+x+1\)

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

\(C=-\left(x^2-2x.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}-1\right)\)

\(C=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{5}{4}\)

Vì \(-\left(x-\dfrac{1}{2}\right)^2\le0\) với mọi x

\(\Rightarrow-\left(x-\dfrac{1}{2}\right)^2+\dfrac{5}{4}\le\dfrac{5}{4}\)

\(\Rightarrow Cmax=\dfrac{5}{4}\Leftrightarrow x=\dfrac{1}{2}\)

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

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

\(D=-2\left(x^2-2.x\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{9}{16}-\dfrac{1}{2}\right)\)

\(D=-2\left(x-\dfrac{3}{4}\right)^2+\dfrac{17}{8}\)

Vì \(-2\left(x-\dfrac{3}{4}\right)^2\le0\) với mọi x

\(\Rightarrow-2\left(x-\dfrac{3}{4}\right)^2+\dfrac{17}{8}\le\dfrac{17}{8}\)

\(\Rightarrow Dmax=\dfrac{17}{8}\Leftrightarrow x=\dfrac{3}{4}\)

\(E=-25x^2-10x+7\)

\(E=-\left(25x^2+10x-7\right)\)

\(E=-\left[\left(5x\right)^2+2.5x+1-8\right]\)

\(E=-\left(5x+1\right)^2+8\)

Vì \(-\left(5x+1\right)^2\le0\) với mọi x

\(\Rightarrow-\left(5x+1\right)^2+8\le8\)

\(\Rightarrow Emax=8\Leftrightarrow5x+1=0\Rightarrow x=-\dfrac{1}{5}\)

Bài 2:

\(A=9x^2+6x+4\)

\(A=\left(3x\right)^2+2.3x+1+3\)

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

Vì \(\left(3x+1\right)^2\ge0\) với mọi x

\(\Rightarrow\left(3x+1\right)^2+3\ge3\)

\(\Rightarrow Amin=3\Leftrightarrow x=-\dfrac{1}{3}\)

\(B=4x^2+4x+12\)

\(B=\left(2x\right)^2+2.2x+1+11\)

\(B=\left(2x+1\right)^2+11\)

Vì \(\left(2x+1\right)^2\ge0\) với mọi x

\(\Rightarrow\left(2x+1\right)^2+11\ge11\)

\(\Rightarrow Bmin=11\Leftrightarrow x=-\dfrac{1}{2}\)

\(C=x^2+x+3\)

\(C=x^2+2x.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}+3\)

\(C=\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}\)

Vì \(\left(x+\dfrac{1}{2}\right)^2\ge0\) với mọi x

\(\Rightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\)

\(\Rightarrow Cmin=\dfrac{11}{4}\Leftrightarrow x=-\dfrac{1}{2}\)

\(D=2x^2+3x+1\)

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

\(D=2\left(x^2+2.x.\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{9}{16}+\dfrac{1}{2}\right)\)

\(D=2\left(x+\dfrac{3}{4}\right)^2-\dfrac{1}{8}\)

Vì \(2\left(x+\dfrac{3}{4}\right)^2\ge0\) với mọi x

\(\Rightarrow2\left(x+\dfrac{3}{4}\right)^2-\dfrac{1}{8}\ge-\dfrac{1}{8}\)

\(\Rightarrow Dmin=-\dfrac{1}{8}\Leftrightarrow x=-\dfrac{3}{4}\)

\(E=64x^2+16x+3\)

\(E=\left(8x\right)^2+2.8x+1+2\)

\(E=\left(8x+1\right)^2+2\)

Vì \(\left(8x+1\right)^2\ge0\) với mọi x

\(\Rightarrow\left(8x+1\right)^2+2\ge2\)

\(\Rightarrow Emin=2\Leftrightarrow x=-\dfrac{1}{8}\)

a: 17-2x=9

=>2x=17-9=8

=>x=8/2=4

b: \(145-135\left(x-2\right)^2=10\)

=>\(135\cdot\left(x-2\right)^2=135\)

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

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

c: \(x\inƯ\left(36\right)\)

=>\(x\in\left\{1;-1;2;-2;3;-3;4;-4;6;-6;9;-9;12;-12;18;-18;36;-36\right\}\)

mà x>12

nên \(x\in\left\{18;36\right\}\)

d: \(x-1\in B\left(9\right)\)

=>\(x-1\in\left\{0;9;18;27;36;45;54;...\right\}\)

=>\(x\in\left\{1;10;19;28;37;46;55;...\right\}\)

mà 25<x<50

nên \(x\in\left\{28;37;46\right\}\)

a)4/3:x=-4/7                         b)5/3.x=7/12

⇒x=-16/21                             ⇒x=7/20

a. \(\dfrac{5}{7}+\dfrac{4}{3}:x=\dfrac{1}{7}\)

<=> \(\dfrac{5}{7}+\dfrac{4}{3}.\dfrac{1}{x}=\dfrac{1}{7}\)

<=> \(\dfrac{5}{7}+\dfrac{4}{3x}=\dfrac{1}{7}\)         ĐKXĐ: x \(\ne\) 0

<=> \(\dfrac{15x}{21x}+\dfrac{28}{21x}=\dfrac{3x}{21x}\)

<=> 15x + 28 = 3x

<=> 15x - 3x = -28

<=> 12x = -28

<=> x = \(\dfrac{-28}{12}=-\dfrac{7}{3}\)

b. \(\dfrac{5}{3}x.\dfrac{-1}{4}=\dfrac{2}{6}\)

<=> \(\dfrac{-5x}{12}=\dfrac{2}{6}\)

<=> -5x . 6 = 12 . 2

<=> -30x = 24

<=> x = \(-\dfrac{4}{5}\)

d: ta có: \(x^2-4x+4=9\left(x-2\right)\)

\(\Leftrightarrow\left(x-2\right)\left(x-11\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=11\end{matrix}\right.\)