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NV
23 tháng 10 2019

a/ ĐKXĐ: ...

\(\Leftrightarrow2\left(x^2-5x-6\right)+\sqrt{x^2-5x-6}-3=0\)

Đặt \(\sqrt{x^2-5x-6}=a\ge0\)

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

\(\Rightarrow\sqrt{x^2-5x-6}=1\Leftrightarrow x^2-5x-7=0\)

b/ ĐKXĐ: ...

\(\Leftrightarrow5\sqrt{3x^2-4x-2}-2\left(3x^2-4x-2\right)+3=0\)

Đặt \(\sqrt{3x^2-4x-2}=a\ge0\)

\(-2a^2+5a+3=0\) \(\Rightarrow\left[{}\begin{matrix}a=3\\a=-\frac{1}{2}\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{3x^2-4x-2}=3\Leftrightarrow3x^2-4x-11=0\)

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

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

\(\frac{a^2-3}{2}-6+a=0\Leftrightarrow a^2+2a-15=0\Rightarrow\left[{}\begin{matrix}x=3\\x=-5\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{2x^2+4x+3}=3\Leftrightarrow2x^2+4x-6=0\)

NV
23 tháng 10 2019

d/ ĐKXĐ: ...

Đặt \(\sqrt{\frac{3x-1}{x}}=a>0\)

\(2a=\frac{1}{a^2}+1\Leftrightarrow2a^3-a^2-1=0\)

\(\Leftrightarrow\left(a-1\right)\left(2a^2+a+1\right)=0\)

\(\Rightarrow a=1\Rightarrow\sqrt{\frac{3x-1}{x}}=1\Leftrightarrow3x-1=x\)

e/ĐKXĐ: ...

\(\Leftrightarrow2\sqrt{\frac{6x-1}{x}}=\frac{x}{6x-1}+1\)

Đặt \(\sqrt{\frac{6x-1}{x}}=a>0\)

\(2a=\frac{1}{a^2}+1\Leftrightarrow2a^3-a^2-1=0\Leftrightarrow\left(a-1\right)\left(2a^2+a+1\right)=0\)

\(\Rightarrow a=1\Rightarrow\sqrt{\frac{6x-1}{x}}=1\Rightarrow6x-1=x\)

f/ ĐKXĐ: ...

Đặt \(\sqrt{\frac{x}{2x-1}}=a>0\)

\(\frac{1}{a}+1+a=3a^2\)

\(\Leftrightarrow3a^3-a^2-a-1=0\)

\(\Leftrightarrow\left(a-1\right)\left(3a^2+2a+1\right)=0\)

\(\Leftrightarrow a=1\Rightarrow\sqrt{\frac{x}{2x-1}}=1\Rightarrow x=2x-1\)

9 tháng 2 2020

A = \(\frac{3x}{2}+\frac{2}{x-1}=3.\frac{x-1}{2}+\frac{2}{x-1}+\frac{3}{2}\)\(\ge2\sqrt{3}+\frac{3}{2}\)

\(\Rightarrow\)min A = \(2\sqrt{3}+\frac{3}{2}\Leftrightarrow x=\frac{2}{\sqrt{3}}+1\)(thỏa mãn)

B = \(x+\frac{3}{3x-1}=\frac{1}{3}\left(3x-1+\frac{9}{3x-1}+1\right)\)\(\ge\frac{1}{3}\left(2\sqrt{9}+1\right)=\frac{7}{3}\)

\(\Rightarrow\)min B = \(\frac{7}{3}\Leftrightarrow x=\frac{4}{3}\)

9 tháng 2 2020

\(A\) \(=\) \(3x^2\left(8-x^2\right)\le3\frac{\left(x^2+8-x^2\right)^2}{4}=48\)

\(\Rightarrow\) maxA = 48 \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)(thỏa mãn)

\(B=\) \(4x\left(8-5x\right)\)\(=\frac{4}{5}.5x\left(8-5x\right)\le\frac{4}{5}.\frac{\left(5x+8-5x\right)^2}{4}=\frac{64}{5}\)

\(\Rightarrow\)max B = \(\frac{64}{5}\Leftrightarrow x=\frac{4}{5}\)(thỏa mãn)

NV
27 tháng 10 2019

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

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

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

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

NV
27 tháng 10 2019

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: ...

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

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

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

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

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

NV
26 tháng 11 2019

a/ ĐKXĐ: \(-\frac{3}{2}\le x\le4\)

\(\sqrt{2x+3}+\sqrt{4-x}=6x-3\left(x+7-2\sqrt{\left(2x+3\right)\left(4-x\right)}\right)-10\)

\(\Leftrightarrow\sqrt{2x+3}+\sqrt{4-x}=3\left(x+7+2\sqrt{\left(2x+3\right)\left(4-x\right)}\right)-52\)

Đặt \(\sqrt{2x+3}+\sqrt{4-x}=a>0\Rightarrow a^2=x+7+2\sqrt{\left(2x+3\right)\left(4-x\right)}\)

Phương trình trở thành:

\(a=3a^2-52\Leftrightarrow3a^2-a-52=0\Rightarrow\left[{}\begin{matrix}a=-4\left(l\right)\\a=\frac{13}{3}\end{matrix}\right.\)

\(\sqrt{2x+3}+\sqrt{4-x}=\frac{13}{3}\)

Phương trình này vô nghiệm nên ko muốn giải tiếp, bạn bình phương lên và chuyển vế thôi :(

b/ ĐKXĐ: \(-\frac{1}{4}\le x\le1\)

Đặt \(\sqrt{4x+1}+2\sqrt{1-x}=a>0\Rightarrow a^2=5+4\sqrt{-4x^2+3x+1}\)

\(\Rightarrow\sqrt{-4x^2+3x+1}=\frac{a^2-5}{4}\)

Pt trở thành:

\(a+10\left(\frac{a^2-5}{4}\right)=13\)

\(\Leftrightarrow5a^2+2a-51=0\Rightarrow\left[{}\begin{matrix}a=3\\a=-\frac{17}{5}\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{-4x^2+3x+1}=\frac{a^2-5}{4}=1\)

\(\Leftrightarrow-4x^2+3x=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\frac{3}{4}\end{matrix}\right.\)

NV
26 tháng 11 2019

c/ \(\Leftrightarrow x^2\left(x^2+2\right)=12-x\sqrt{2x^2+4}\)

\(\Leftrightarrow x^2\left(2x^2+4\right)=24-2x\sqrt{2x^2+4}\)

Đặt \(x\sqrt{2x^2+4}=a\) ta được:

\(a^2=24-2a\Leftrightarrow a^2+2a-24=0\Leftrightarrow\left[{}\begin{matrix}a=4\\a=-6\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x\sqrt{2x^2+4}=4\left(x>0\right)\\x\sqrt{2x^2+4}=-6\left(x< 0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2\left(2x^2+4\right)=16\\x^2\left(2x^2+4\right)=36\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^4+2x^2-8=0\\x^4+2x^2-18=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2=2\\x^2=-4\left(l\right)\\x^2=\sqrt{19}-1\\x^2=-\sqrt{19}-1\left(l\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}< 0\left(l\right)\\x=-\sqrt{\sqrt{19}-1}\\x=\sqrt{\sqrt{19}-1}>0\left(l\right)\end{matrix}\right.\)

1 tháng 10 2019

ĐK: \(x^4-4x^3+14x-11\ge0\) (*)

\(PT\Leftrightarrow\left\{{}\begin{matrix}x\le1\\x^4-4x^3+14x-11=x^2-2x+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le1\\x^4-4x^3-x^2+16x-12=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le1\\\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x+2\right)=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)(tm)

NV
1 tháng 10 2019

e/ ĐKXĐ: \(x\ge1\)

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

\(\Leftrightarrow\sqrt{x-1}=x-1\)

\(\Leftrightarrow x-1=x^2-2x+1\)

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

f/ \(\Leftrightarrow\left\{{}\begin{matrix}x+5\ge0\\x^3+x^2+6x+28=\left(x+5\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-5\\x^3-4x+3=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-5\\\left(x-1\right)\left(x^2+x-3\right)=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=\frac{-1\pm\sqrt{13}}{2}\\\end{matrix}\right.\)

13 tháng 12 2019

a, ĐK: \(6x^2-12x+7\ge0\) (*)

\(PT\Leftrightarrow\left\{{}\begin{matrix}x^2-2x\ge0\\6x^2-12x+7=x^4-4x^3+4x^2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x^2-2x\ge0\\x^4-4x^3-2x^2+12x-7=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2-2x\ge0\\\left(x-1\right)^2\left(x^2-2x-7\right)=0\end{matrix}\right.\) \(\Rightarrow x=1\pm2\sqrt{2}\) (thỏa mãn ĐK)

Vậy...

NV
16 tháng 8 2020

8.

ĐKXĐ: \(x\ge\frac{2}{3}\)

\(\Leftrightarrow\frac{9\left(x+3\right)}{\sqrt{4x+1}+\sqrt{3x-2}}=x+3\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\left(l\right)\\\frac{9}{\sqrt{4x+1}+\sqrt{3x-2}}=1\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\sqrt{4x+1}+\sqrt{3x-2}=9\)

\(\Leftrightarrow\sqrt{4x+1}-5+\sqrt{3x-2}-4=0\)

\(\Leftrightarrow\frac{4\left(x-6\right)}{\sqrt{4x+1}+5}+\frac{3\left(x-6\right)}{\sqrt{3x-2}+4}=0\)

\(\Leftrightarrow\left(x-6\right)\left(\frac{4}{\sqrt{4x+1}+5}+\frac{3}{\sqrt{3x-2}+4}\right)=0\)

\(\Leftrightarrow x=6\)

NV
16 tháng 8 2020

6.

ĐKXD: ...

\(\Leftrightarrow2\left(x^2-6x+9\right)+\left(x+5-4\sqrt{x+1}\right)=0\)

\(\Leftrightarrow2\left(x-3\right)^2+\frac{\left(x-3\right)^2}{x+5+4\sqrt{x+1}}=0\)

\(\Leftrightarrow\left(x-3\right)^2\left(2+\frac{1}{x+5+4\sqrt{x+1}}\right)=0\)

\(\Leftrightarrow x=3\)

7.

\(\sqrt{x-\frac{1}{x}}-\sqrt{2x-\frac{5}{x}}+\frac{4}{x}-x=0\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x-\frac{1}{x}}=a\ge0\\\sqrt{2x-\frac{5}{x}}=b\ge0\end{matrix}\right.\) \(\Rightarrow a^2-b^2=\frac{4}{x}-x\)

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

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

\(\Leftrightarrow a=b\Leftrightarrow x-\frac{1}{x}=2x-\frac{5}{x}\)

\(\Leftrightarrow x=\frac{4}{x}\Rightarrow x=\pm2\)

Thế nghiệm lại pt ban đầu để thử (hoặc là bạn tìm ĐKXĐ từ đầu)