Question

Giải phương trình vô tỉ:

a) \(4x^2-4x-10=\sqrt{8x^2-6x-10}\)

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

c) \(\sqrt{3x+8+6\sqrt{3x-1}}+\sqrt{3x+8-6\sqrt{3x-1}}=3x+4\)

d) \(2x\sqrt{x^2-x+1}+4\sqrt{3x+1}=2x^2+2x+6\)

Answer 1

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Answer 2

cho đúng nha

Answer 3

a,Ta có:\(4x^2-4x-10=\sqrt{8x^2-6x-5}\)

\(\Leftrightarrow16x^4+16x^2+100-80x^2-32x^3+80x=8x^2-6x-5\)

\(\Leftrightarrow16x^4-32x^3-64x^2+80x+100-8x^2+6x+5=0\)

\(\Leftrightarrow16x^4-32x^3-72x^2+86x+110=0\)

\(\Leftrightarrow2\left(x+1\right)\left(2x-5\right)\left(4x^2-2x-11\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{5}{2}\\4x^2-2x-11=0\Rightarrow\left[{}\begin{matrix}\dfrac{1+3\sqrt{5}}{4}\\\dfrac{1-3\sqrt{5}}{4}\end{matrix}\right.\end{matrix}\right.\)

Vậy ....

Answer 4

a, Tớ thiếu ĐKXĐ \(x\ge\dfrac{\sqrt{11}+1}{2}\)

\(\Rightarrow\) Chỉ lấy giá trị x=\(\dfrac{5}{2}\) thôi nhé.