Question

[Đinh Doãn Nam]

a)\(\sqrt{1-x}\left(x-3x^2\right)=x^3-3x^2+2x+6\)

b)\(x^2+x+12\sqrt{x+1}=36\)

c)\(3x-1+\frac{x-1}{4x}=\sqrt{3x+1}\)

d)\(\sqrt{x^2+12}-3x=\sqrt{x^2+5}-5\)

e)\(4x^2+12+\sqrt{x-1}=4\left(x\sqrt{5x-1}+\sqrt{9-5x}\right)\)

f)\(4x^3-25x^2+43x+x\sqrt{3x-2}=22+\sqrt{3x-2}\)

g)\(2\left(x+1\right)\sqrt{x}+\sqrt{3\left(2x^3+5x^2+4x+1\right)}=5x^3-3x^2+8\)

h)\(\sqrt{x^2+12}-\sqrt{x^2+5}=3x-5\)

i)\(\sqrt{1-3x}-\sqrt[3]{3x-1}=\left|6x-2\right|\)

k)\(\sqrt{2x^3+3x^2-1}=2x^2+2x-x^3-1\)

l)\(\sqrt{x^2+x-2}+x^2=\sqrt{2\left(x-1\right)}+1\)

Answer 1

[Huyền]

2,\(pt\Leftrightarrow12\left(\sqrt{x+1}-2\right)+x^2+x-12=0\)

\(\Leftrightarrow12\cdot\frac{x-3}{\sqrt{x+1}+2}+\left(x-3\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(\frac{12}{\sqrt{x+1}+2}+x+4\right)=0\)

Vì \(\left(\frac{12}{\sqrt{x+1}+2}+x+4\right)\ge0\left(\forall x>-1\right)\)

\(\Rightarrow x=3\)

Answer 2

[Huyền]

c,\(pt\Leftrightarrow3\left(x-1\right)+\frac{x-1}{4x}+\left(2-\sqrt{3x+1}\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(3+\frac{1}{4x}+\frac{1}{2+\sqrt{3x+1}}\right)=0\)

\(\Rightarrow x=1\)

\(3+\frac{1}{4x}+\frac{1}{2+\sqrt{3x+1}}=0\)

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