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mik ko biết vì mới chỉ học lớp 6

ĐKXĐ: \(x\ge\frac{1}{2}\)

Đề \(\Rightarrow\sqrt{\frac{x+7}{x+1}}-\sqrt{3}+8-2x^2-\left(\sqrt{2x-1}-\sqrt{3}\right)=0\)

Nhân liên hợp ta được:

\(\frac{\left(\sqrt{\frac{x+7}{x+1}}-\sqrt{3}\right)\left(\sqrt{\frac{x+7}{x+1}}+\sqrt{3}\right)}{\sqrt{\frac{x+7}{x+1}}+\sqrt{3}}+2\left(4-x^2\right)-\frac{\left(\sqrt{2x-1}-\sqrt{3}\right)\left(\sqrt{2x+1}+\sqrt{3}\right)}{\sqrt{2x+1}+\sqrt{3}}=0\)

\(\Rightarrow\frac{\frac{x+7}{x+1}-3}{\sqrt{\frac{x+7}{x+1}}+\sqrt{3}}+2\left(4-x^2\right)-\frac{2x-1-3}{\sqrt{2x+1}+\sqrt{3}}=0\)

\(\Rightarrow\frac{\frac{-2x+4}{x+1}}{\sqrt{\frac{x+7}{x+1}}+\sqrt{3}}+2\left(2-x\right)\left(2+x\right)-\frac{2x-4}{\sqrt{2x+1}+\sqrt{3}}=0\)

\(\Rightarrow\left(x-2\right)\left[\frac{-2}{\left(x+1\right)\left(\sqrt{\frac{x+7}{x+1}}+\sqrt{3}\right)}-2\left(2+x\right)-\frac{2}{\sqrt{2x+1}+\sqrt{3}}\right]=0\)

mà \(-\frac{2}{\left(x+1\right)\left(\sqrt{\frac{x+7}{x+1}}+\sqrt{3}\right)}-2\left(2+x\right)-\frac{2}{\sqrt{2x+1}+\sqrt{3}}< 0\)

=> x - 2 = 0 => x = 2

                                                   Vậy x = 2

rảnh  quá 

Tham khảo:

Câu hỏi của Huyen123 Đaothi - Toán lớp 10 | Học trực tuyến

TXĐ: \(x\ge0\)

Phương trình đã cho tương đương:

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

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

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

\(\Leftrightarrow x-1=0\) (do \(1+\dfrac{1}{\sqrt{2x+1}+\sqrt{3x}}>0\) \(\forall x\ge0\))

\(\Leftrightarrow x=1\)

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

Điều kiện : x\(\ge0\)

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

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

\(\Leftrightarrow2x+1=\left(x-1\right)^2+2\left(x-1\right)\sqrt{3x}+3x\)

\(\Leftrightarrow2x+1=x^2-2x+1+2\left(x-1\right)\sqrt{3x}+3x\)

\(\Leftrightarrow2x+1-x^2-x-x-2\left(x-1\right)\sqrt{3x}=0\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\-x-2\sqrt[]{3x}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\-\sqrt{x}\left(\sqrt{x}+2\sqrt{3}\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\-\sqrt{x}=0\\\sqrt{x}+2\sqrt{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=0\\\sqrt{x}=-2\sqrt{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=0\\x\in\varnothing\end{matrix}\right.\) Vậy pt tập nghiệm S={1;0}

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

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

\(\Leftrightarrow2x+2\sqrt{x^2-2x+1}=2\Leftrightarrow2\left(x+\sqrt{\left(x-1\right)^2}\right)=2\Leftrightarrow x+|^{ }_{ }x-1|=1\)

\(\Leftrightarrow|^{ }_{ }x-1|^{ }_{ }=1-x\Leftrightarrow x-1< 0\Leftrightarrow x< 1\)

vậy x<1

mik hỏi bạn mik bảo x<10

duphuongthao làm đúng rồi, tham khảo nha !

Đặt \(\sqrt{x+1}=a\)                                            \(ĐKXĐ:x\ge0\)

        \(\sqrt{3x}=b\)                                               

Ta có: \(a-b=b^2-a^2\)

\(\Leftrightarrow a-b+a^2-b^2=0\)

\(\Leftrightarrow\left(a-b\right)+\left(a+b\right)\left(a-b\right)=0\)

\(\Leftrightarrow\left(a-b\right)\left(a+b+1\right)=0\)

Mà \(a+b+1>0\forall x\)

\(\Rightarrow a-b=0\)

\(\Leftrightarrow a=b\)

\(\Leftrightarrow\sqrt{x+1}=\sqrt{3x}\)

\(\Leftrightarrow x+1=3x\)

\(\Leftrightarrow2x-1=0\)

\(\Leftrightarrow x=\frac{1}{2}\)

Vậy phương trình có tập nghiệm \(S=\left\{\frac{1}{2}\right\}\)

\(ĐKXĐ:x\ge0\)

Ta có PT \(\Leftrightarrow\sqrt{x+1}-\sqrt{3x}-\left(2x-1\right)=0\)

\(\Leftrightarrow\left(\sqrt{x+1}-\frac{\sqrt{6}}{2}\right)-\left(\sqrt{3x}-\frac{\sqrt{6}}{2}\right)-\left(2x-1\right)=0\)

\(\Leftrightarrow\frac{x+1-\frac{6}{4}}{\sqrt{x+1}+\frac{\sqrt{6}}{2}}-\frac{3x-\frac{6}{4}}{\sqrt{3x}+\frac{\sqrt{6}}{2}}-\left(2x-1\right)=0\)

\(\Leftrightarrow\frac{x-\frac{1}{2}}{\sqrt{x+1}+\frac{\sqrt{6}}{2}}-\frac{3\left(x-\frac{1}{2}\right)}{\sqrt{3x}+\frac{\sqrt{6}}{2}}-2\left(x-\frac{1}{2}\right)=0\)

\(\Leftrightarrow\left(x-\frac{1}{2}\right)\left(\frac{1}{\sqrt{x+1}+\frac{\sqrt{6}}{2}}-\frac{3}{\sqrt{3x}+\frac{\sqrt{6}}{2}}-2\right)=0\)

\(\Rightarrow x=\frac{1}{2}\)(TMĐKXĐ)