a) \(4\sqrt{2x+1}-\sqrt{8x+4}+\dfrac{1}{2}\sqrt{32x+16}=12\) (ĐK: \(x\ge-\dfrac{1}{2}\)) 

\(\Leftrightarrow4\sqrt{2x+1}-\sqrt{4\left(2x+1\right)}+\dfrac{1}{2}\cdot4\sqrt{2x+1}=12\)

\(\Leftrightarrow4\sqrt{2x+1}-2\sqrt{2x+1}+2\sqrt{2x+1}=12\)

\(\Leftrightarrow4\sqrt{2x+1}=12\)

\(\Leftrightarrow\sqrt{2x+1}=\dfrac{12}{4}\)

\(\Leftrightarrow2x+1=3^2\)

\(\Leftrightarrow2x=9-1\)

\(\Leftrightarrow2x=8\)

\(\Leftrightarrow x=\dfrac{8}{2}\)

\(\Leftrightarrow x=4\left(tm\right)\)

b) \(\sqrt{4x^2-4x+1}=5\)

\(\Leftrightarrow\sqrt{\left(2x-1\right)^2}=5\)

\(\Leftrightarrow\left|2x-1\right|=5\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=5\left(x\ge\dfrac{1}{2}\right)\\2x-1=-5\left(x< \dfrac{1}{2}\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{6}{2}\\x=-\dfrac{4}{2}\end{matrix}\right.\)

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

c) \(\dfrac{2\sqrt{x}-3}{\sqrt{x}-1}=-\dfrac{1}{2}\)(ĐK: \(x\ge0;x\ne1\))

\(\Leftrightarrow-\left(\sqrt{x}-1\right)=2\left(2\sqrt{x}-3\right)\)

\(\Leftrightarrow-\sqrt{x}+1=4\sqrt{x}-6\)

\(\Leftrightarrow4\sqrt{x}+\sqrt{x}=1+6\)

\(\Leftrightarrow5\sqrt{x}=7\)

\(\Leftrightarrow\sqrt{x}=\dfrac{7}{5}\)

\(\Leftrightarrow x=\dfrac{49}{25}\left(tm\right)\)

a: \(\left(2x-3\right)\left(3x^2+1\right)-6x\left(x^2-x+1\right)+3x^2-2x=10\)

\(\Leftrightarrow6x^3+2x-9x^2-3-6x^3+6x^2-6x+3x^2-2x=10\)

\(\Leftrightarrow-6x-3=10\)

=>-6x=13

hay x=-13/6

b: \(\Leftrightarrow3x^2-3x+x-2-3x^2+5x=-8-5x\)

=>3x-2=-5x-8

=>8x=-6

hay x=-3/4

c: \(\Leftrightarrow64x^3-27-64x^3+32x^2-32x^2+x=20\)

=>x-27=20

hay x=47

      \(27x^3-27x^2+18x-4\)

\(=27x^3-9x^2-18x^2+6x+12x-4\)

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

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

       \(x^2+2xy+y^2-x-y-12\)

\(=\left(x+y\right)^2-\left(x+y\right)-12\)

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

\(=\left(x+y\right)\left(x+y-4\right)+3\left(x+y-4\right)=\left(x+y+3\right)\left(x+y-4\right)\)  \(P=\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

   \(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\) (nhóm 2 cái đầu với cuối lại với nhau, 2 cái giữa vào 1 nhóm)

Đặt \(x^2+7x+11=a\)

Ta có: \(P=\left(a-1\right)\left(a+1\right)-24\)

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

\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)

d,   \(4x^4-32x^2+1\)

\(=4x^4+4x^2+1-36x^2\)

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

huyển vế: 
(x-2)(x-6)(x-3)(x-4)- 72X^2 

(x-2)(x-6) 
= (x^2 - ... +12) 
số giữa: 
-6x -2x = -8x 

(x-3)(x-4) 
= (x^2 ... +12) 
số giữa: 
-4x -3x = -7x 

nhân 2 số giữa với nhau: 
(-8x)(-7x) = +56x^2 
-72x^2 +56x^2 = -16x^2 = (-16x)(x) 

Đáp số: 
(x^2 -16x +12)(x^2 +x +12)

1) \(x^2+2xy+y^2-x-y-12\)

= \(\left(x+y\right)^2-\left(x+y\right)-12\)

Đặt \(x+y=z\) (đặt ẩn phụ)

\(\Rightarrow z^2-z-12\)

\(=z^2+3z-4z-12\)

\(=z\left(z+3\right)-4\left(z+3\right)\)

\(=\left(z+3\right)\left(z-4\right)\)

Khi đó: \(\left(x+y+3\right)\left(x+y-4\right)\)

#HuyenAnh

1/ \(x^2+2xy+y^2-x-y-12=\left(x+y\right)^2+6\left(x+y\right)+9-7\left(x+y\right)-21\)

\(=\left(x+y+3\right)^2-7\left(x+y+3\right)=\left(x+y+3\right)\left(x+y-4\right)\)

2/ \(4x^4-32x^2+1=\left(4x^4+4x^2+1\right)-36x^2\)

\(=\left(2x^2+1\right)^2-36x^2=\left(2x^2-6x+1\right)\left(2x^2+6x+1\right)\)

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

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

Còn lại tự làm nhé

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coá nhoa bn