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.

Giải hệ đầu tiên:

\(\left\{{}\begin{matrix}4x^2y-xy^2=5\\64x^3-y^3=61\left(1\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}xy\left(4x-y\right)=5\\\left(4x-y\right)\left(16x^2+4xy+y^2\right)=61\end{matrix}\right.\)

\(\Leftrightarrow5\left(4x-y\right)\left(16x^2+4xy+y^2\right)-61xy\left(4x-y\right)=0\)

Hiển nhiên \(4x-y\ne0\) nên ta chia cả 2 vế cho \(\left(4x-y\right)\)

\(\Leftrightarrow80x^2-41xy+5y^2=0\)

\(\Leftrightarrow\left(16x-5y\right)\left(5x-y\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y=\frac{16}{5}x\\y=5x\end{matrix}\right.\) Lần lượt thay vào (1) để tìm x.

Từ phương trình chứa căn ban đầu ta có: ĐKXĐ là \(-\frac{11}{5}\le x\le6\)

\(\sqrt{5x+11}-6+1-\sqrt{6-x}+5x^2-14x-55=0\)

\(\Leftrightarrow\frac{5\left(x-5\right)}{\sqrt{5x+11}+6}+\frac{x-5}{\sqrt{6-x}+1}+\left(x-5\right)\left(5x+11\right)=0\) (1)

Dễ thấy có nghiệm \(x=5\), thử lại thỏa mãn.

Với \(x\ne5\), chia cả 2 vế cho \(\left(x-5\right)\)

\(\Leftrightarrow\frac{5}{\sqrt{5x+11}+6}+\frac{1}{\sqrt{6-x}+1}+5x+11=0\) (2)

Vế trái của (2) luôn lớn hơn 0 với mọi \(x\ge\frac{-11}{5}\)

Vậy \(x=5\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

a, \(\sqrt{\left(2x-3\right)^2}=7\\ \Rightarrow\left|2x-3\right|=7\\ \Rightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

c, \(\sqrt{x^2-9}-3\sqrt{x-3}=0\\ \Rightarrow\sqrt{x-3}\sqrt{x+3}-3\sqrt{x-3}=0\\ \Rightarrow\sqrt{x-3}\left(\sqrt{x+3}-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\\sqrt{x+3}-3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x-3=0\\x+3=9\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=6\left(tm\right)\end{matrix}\right.\)

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 

\(a,3x-4y-3y+4x\)

\(=3\left(x-y\right)+4\left(x-y\right)\)

\(=\left(3+4\right)\left(x-y\right)=7\left(x-y\right)\)

\(b,\left(a^3+2ab+b^2\right)-\left(a^3+b^3\right)\)

\(=a^3+2ab+b^2-a^3-b^3\)

\(=2ab+b^2-b^3\)

\(=b\left(2a+b-b^2\right)\)

\(c,48b^3-24b^2=3b\)

\(48b^3-24b^2-3b=0\)

\(b\left(48b^2-24b-3\right)=0\)

Answer:

\(3x^2-4x=0\)

\(\Rightarrow x\left(3x-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{4}{3}\end{cases}}\)

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

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

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

\(x^2-5x+6=0\)

\(\Rightarrow x^2-2x-3x+6=0\)

\(\Rightarrow\left(x^2-2x\right)-\left(3x-6\right)=0\)

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

\(\Rightarrow\orbr{\begin{cases}x-2=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}\)

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

\(\Rightarrow5x\left(x-3\right)-\left(x-3\right)=0\)

\(\Rightarrow\left(5x-1\right)\left(x-3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}5x-1=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=3\end{cases}}\)

\(x^2-2x+5=0\)

\(\Rightarrow\left(x^2-2x+1\right)+4=0\)

\(\Rightarrow\left(x-1\right)^2=-4\) (Vô lý)

Vậy không có giá trị \(x\) thoả mãn

\(x^2+x-6=0\)

\(\Rightarrow x^2+3x-2x-6=0\)

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

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

\(\Rightarrow\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}}\)

a) \(\Rightarrow x^2\left(x^2-64\right)=0\Rightarrow x^2\left(x-8\right)\left(x+8\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=8\\x=-8\end{matrix}\right.\)

b) \(\Rightarrow x^2\left(x-6\right)+3x\left(x-6\right)+21\left(x-6\right)=0\Rightarrow\left(x-6\right)\left(x^2+3x+21\right)=0\)

\(\Rightarrow x=6\)

a: Ta có: \(x\left(x-3\right)-x^2+5=0\)

\(\Leftrightarrow-3x+5=0\)

hay \(x=\dfrac{5}{3}\)

b: Ta có: \(x^2-6x=0\)

\(\Leftrightarrow x\left(x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)