em ms hok lớp 1

a/ \(x+\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}}=4\)

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

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

Làm nốt

b/ \(\sqrt{2x+4-6\sqrt{2x-5}}+\sqrt{2x-4+2\sqrt{2x-5}}=4\)

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

Làm nốt

a/ ĐKXĐ: ...

\(\Leftrightarrow3\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)-7\)

Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=a>0\Rightarrow a^2=x+\frac{1}{4x}+1\)

\(\Rightarrow x+\frac{1}{4x}=a^2-1\)

Pt trở thành:

\(3a=2\left(a^2-1\right)-7\)

\(\Leftrightarrow2a^2-3a-9=9\Rightarrow\left[{}\begin{matrix}a=3\\a=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)

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

\(\Leftrightarrow2x-6\sqrt{x}+1=0\)

\(\Rightarrow\sqrt{x}=\frac{3+\sqrt{7}}{2}\Rightarrow x=\frac{8+3\sqrt{7}}{2}\)

b/ ĐKXĐ:

\(\Leftrightarrow5\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)+4\)

Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=a>0\Rightarrow x+\frac{1}{4x}=a^2-1\)

\(\Rightarrow5a=2\left(a^2-1\right)+4\Leftrightarrow2a^2-5a+2=0\)

\(\Rightarrow\left[{}\begin{matrix}a=2\\a=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{x}+\frac{1}{2\sqrt{x}}=2\\\sqrt{x}+\frac{1}{2\sqrt{x}}=\frac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x-4\sqrt{x}+1=0\\2x-\sqrt{x}+1=0\left(vn\right)\end{matrix}\right.\)

c/ ĐKXĐ: ...

\(\Leftrightarrow\sqrt{2x^2+8x+5}-4\sqrt{x}+\sqrt{2x^2-4x+5}-2\sqrt{x}=0\)

\(\Leftrightarrow\frac{2x^2-8x+5}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\frac{2x^2-8x+5}{\sqrt{2x^2-4x+5}+2\sqrt{x}}=0\)

\(\Leftrightarrow\left(2x^2-8x+5\right)\left(\frac{1}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\frac{1}{\sqrt{2x^2-4x+5}+2\sqrt{x}}\right)=0\)

\(\Leftrightarrow2x^2-8x+5=0\)

d/ ĐKXĐ: ...

\(\Leftrightarrow x+1-\frac{15}{6}\sqrt{x}+\sqrt{x^2-4x+1}-\frac{1}{2}\sqrt{x}=0\)

\(\Leftrightarrow\frac{x^2-\frac{17}{4}x+1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{x^2-\frac{17}{4}x+1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}=0\)

\(\Leftrightarrow\left(x^2-\frac{17}{4}x+1\right)\left(\frac{1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}\right)=0\)

\(\Leftrightarrow x^2-\frac{17}{4}x+1=0\)

\(\Leftrightarrow4x^2-17x+4=0\)

e/ ĐKXĐ: ...

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

Nhận thấy \(x=0\) không phải nghiệm, pt tương đương:

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

Đặt \(\sqrt{\frac{x^2-1}{x}}=a\ge0\)

\(a^2+2a=3\Leftrightarrow a^2+2a-3=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-3\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow\sqrt{\frac{x^2-1}{x}}=1\Leftrightarrow x^2-1=x\Leftrightarrow x^2-x-1=0\)

f/ ĐKXĐ: ...

\(\Leftrightarrow x^2-6+x\sqrt{\frac{x^2-6}{x}}-6x=0\)

Nhận thấy \(x=0\) ko phải nghiệm, pt tương đương:

\(\frac{x^2-6}{x}+\sqrt{\frac{x^2-6}{x}}-6=0\)

Đặt \(\sqrt{\frac{x^2-6}{x}}=a\ge0\)

\(a^2+a-6=0\Rightarrow\left[{}\begin{matrix}a=2\\a=-3\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{\frac{x^2-6}{x}}=2\Leftrightarrow x^2-4x-6=0\)

a/ ĐKXĐ: ...

\(\Leftrightarrow\sqrt{2x^2+5x+2}=2\sqrt{2x^2+5x-6}\)

\(\Leftrightarrow2x^2+5x+2=4\left(2x^2+5x-6\right)\)

\(\Leftrightarrow6x^2+15x-26=0\)

b/ ĐKXĐ: ...

Đặt \(\sqrt[5]{\frac{16x}{x-1}}=a\)

\(a+\frac{1}{a}=\frac{5}{2}\Leftrightarrow a^2-\frac{5}{2}a+1=0\)

\(\Rightarrow\left[{}\begin{matrix}a=2\\a=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt[5]{\frac{16x}{x-1}}=2\\\sqrt[5]{\frac{16x}{x-1}}=\frac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}16x=32\left(x-1\right)\\16x=\frac{1}{32}\left(x-1\right)\end{matrix}\right.\)

c/ĐKXĐ: ...

\(\Leftrightarrow x^2-2x-\sqrt{6x^2-12x+7}=0\)

Đặt \(\sqrt{6x^2-12x+7}=a\ge0\Rightarrow x^2-2x=\frac{a^2-7}{6}\)

\(\frac{a^2-7}{6}-a=0\Leftrightarrow a^2-6a-7=0\)

\(\Rightarrow\left[{}\begin{matrix}a=-1\left(l\right)\\a=7\end{matrix}\right.\) \(\Rightarrow\sqrt{6x^2-12x+7}=7\)

\(\Leftrightarrow6x^2-12x-42=0\)

d/ \(\Leftrightarrow x^2+x+4-\sqrt{x^2+x+4}-2=0\)

Đặt \(\sqrt{x^2+x+4}=a>0\)

\(a^2-a-2=0\Rightarrow\left[{}\begin{matrix}a=-1\left(l\right)\\a=2\end{matrix}\right.\)

\(\Rightarrow\sqrt{x^2+x+4}=2\Rightarrow x^2+x=0\)

e/ \(\Leftrightarrow x^2+2x+\sqrt{3x^2+6x+4}-2=0\)

Đặt \(\sqrt{3x^2+6x+4}=a>0\Rightarrow x^2+2x=\frac{a^2-4}{3}\)

\(\frac{a^2-4}{3}+a-2=0\)

\(\Leftrightarrow a^2+3a-10=0\Rightarrow\left[{}\begin{matrix}a=2\\a=-5\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{3x^2+6x+4}=2\Rightarrow3x^2+6x=0\)

ĐKXĐ:...

a/ \(\sqrt{2x^2+5x+2}=1+2\sqrt{2x^2+5x-6}\)

\(\Leftrightarrow2x^2+5x+2=4\left(2x^2+5x-6\right)+1+4\sqrt{2x^2+5x-6}\)

\(\Leftrightarrow3\left(2x^2+6x-6\right)+4\sqrt{2x^2+5x-6}-7=0\)

Đặt \(\sqrt{2x^2+5x-6}=a\ge0\)

\(3a^2+4a-7=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-\frac{7}{3}\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{2x^2+5x-6}=1\)

\(\Leftrightarrow2x^2+5x-7=0\)