\(a,8x-3=5x+12\\ \Leftrightarrow8x-5x=12+3\\ \Leftrightarrow3x=15\\ \Leftrightarrow x=\dfrac{15}{3}=5\)

\(b,x-12+4x=25+2x-1\\ \Leftrightarrow x+4x-2x=25-1+12\\ \Leftrightarrow3x=36\\ \Leftrightarrow x=\dfrac{36}{3}=12\)

\(c,7-\left(2x+4\right)=-\left(x+4\right)\\ \Leftrightarrow7-2x-4=-x-4\\ \Leftrightarrow-2x+x=-4+4-7\\ \Leftrightarrow-x=-7\\ \Leftrightarrow x=7\)

\(d,3-4x\left(45-2x\right)=8x^2+x-300\\ \Leftrightarrow3-100x+8x^2=8x^2+x-300\\ \Leftrightarrow8x^2-8x^2-100x-x=-300-3\\ \Leftrightarrow-101x=-303\\ \Leftrightarrow x=\dfrac{-303}{-101}=3\)

Đề câu d của bạn hình như sai dấu ý

sửa lại

.d. 3 - 4x (25 - 2x) = 8x ² + x - 300

a. \(x^2+4x+4=x^2+2\cdot x\cdot2+2^2=\left(x+2\right)^2\)

b. \(4x^2-4x+1=\left(2x\right)^2-2\cdot2x\cdot1+1^2=\left(2x-1\right)^2\)

c. \(4x^2+12x+9=\left(2x\right)^2+2\cdot2x\cdot3+3^2=\left(2x+3\right)^2\)

d. \(9x^2+30x+25=\left(3x\right)^2+2\cdot3x\cdot5+5^2=\left(3x+5\right)^2\)

e. \(4x^2-20x+25=\left(2x\right)^2-2\cdot2x\cdot5+5^2=\left(2x+5\right)^2\)

\(x^2+4x+4=\left(x+2\right)^2\)

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

\(4x^2+12x+9=\left(2x+3\right)^2\)

\(9x^2+30x+25=\left(3x+5\right)^2\)

\(4x^2-20x+25=\left(2x+5\right)^2\)

tik mik nha

giải

x^2+4x+4

=x^2+2.x.2+2^2

a: ta có: \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\)

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

\(\Leftrightarrow\sqrt{x-1}=1\)

hay x=2

c: Ta có: \(\sqrt{1-2x^2}=x-1\)

\(\Leftrightarrow1-2x^2=x^2-2x+1\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=\dfrac{2}{3}\left(loại\right)\end{matrix}\right.\)

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

n) \(\dfrac{4}{49}-4x^2=\left(\dfrac{2}{7}-2x\right)\left(\dfrac{2}{7}+2x\right)\)

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

Mik cần gấp 

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

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

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

\(x^2-25=\left(x-5\right)\left(x+5\right)\)

mình nghĩ căn đầu tiên phải là `x+1` mới đúng kiểu đề á, còn không phải thì bạn cmt nói mình nha=))

ĐK: \(x\ge-1\)

PT trở thành:

\(\sqrt{x+1}+\sqrt{4}.\sqrt{x+1}-\sqrt{25}.\sqrt{x+1}=-8\\ \Leftrightarrow\sqrt{x+1}+2\sqrt{x+1}-5\sqrt{x+1}=-8\\ \Leftrightarrow\left(1+2-5\right)\sqrt{x+1}=-8\\ \Leftrightarrow-2\sqrt{x+1}=-8\\ \Leftrightarrow\sqrt{x+1}=-\dfrac{8}{-2}=4\\ \Leftrightarrow x+1=4^2=16\\ \Leftrightarrow x=16-1=15\left(tm\right)\)

\(\text{2x - (x - 3)(5 - x) = (x+4)}^2.\)

\(\Leftrightarrow2x-\left(5x-x^2-15+3x\right)=x^2+8x+16.\)

\(\Leftrightarrow2x-5x+x^2+15-3x-x^2-8x-16=0.\)

\(\Leftrightarrow-14x-1=0.\Leftrightarrow x=\dfrac{-1}{14}.\)

\(\text{(4x + 1)(x - 2) + 25 = (2x+3)}^2-4x.\)

\(\Leftrightarrow4x^2-8x+x-2+25=4x^2+12x+9-4x.\)

\(\Leftrightarrow-15x+14=0.\Leftrightarrow x=\dfrac{14}{15}.\)

Câu a: 

TH1: \(x+\sqrt{\left(2x-1\right)^2}=2\Leftrightarrow x+2x-1=2\Leftrightarrow x=1\)

TH2:\(x+\sqrt{\left(2x-1\right)^2}=2\Leftrightarrow x-2x+1=2\Leftrightarrow x=-1\)

ĐK:  \(x\le2\)

\(x+\sqrt{4x^2-4x+1}=2\)

\(\Leftrightarrow\)\(\sqrt{4x^2-4x+1}=2-x\)

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

\(\Leftrightarrow\)\(3x^2=3\)

\(\Leftrightarrow\)\(x=\pm1\)(t/m)

Vậy...

\(1-\sqrt{4x^2-20x+25}=0\)

\(\Leftrightarrow\)\(\sqrt{4x^2-20x+25}=1\)

\(\Leftrightarrow\)\(4x^2-20x+24=0\)

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

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

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=2\\x=3\end{cases}}\)

Vậy...

a) \(\sqrt{4x^2-4x+1}=2-x\left(x\le2\right)\)

<=> \(4x^2-4x+1=\left(2-x\right)^2\)

<=> \(3x^2-3=0\)
<=> \(x_1=1_{ }\)(TM)          ;       \(x_2=-1_{ }_{ }\)(TM)

b)\(1-\sqrt{4x^4-20x^2+25}=0\)

Đặt x² = a ( a\(\ge\)0)

=> pt có dạng :

\(1-\sqrt{4a^2-20a+25}=0\)

<=> \(\sqrt{4a^2-20a+25}=1\)

<=> \(4a^2-20a+25=1\)

=> \(a_1=3\)\(\left(TM\right)\)                  ;      \(a_2=2\left(TM\right)\)

\(a_1=3\Rightarrow x=\pm\sqrt{3}\)

\(a_2=2\Rightarrow x=\pm\sqrt{2}\)

a) Ta có :  2 x : 2 2 = 2 5  nên x = 7.

b) Ta có:  3 x : 3 2 = 3 5  nên x = 7.

c) Ta có :  4 4 : 4 x = 4 2  nên x = 2.

d) Ta có :  5 x : 5 2 = 5 2  nên x = 4,

e) Ta có:  5 x + 1 : 5 = 5 4  nên x = 4.

f) Ta có :  4 2 x - 1 : 4 = 4 2  nên x = 2