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1/ \(3x^2+4x-3=4x\sqrt{4x-3}\)
\(\Leftrightarrow\left(4x^2-4x\sqrt{4x-3}+4x-3\right)-x^2=0\)
\(\Leftrightarrow\left(2x-\sqrt{4x-3}\right)^2-x^2=0\)
\(\Leftrightarrow\left(3x-\sqrt{4x-3}\right)\left(x-\sqrt{4x-3}\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}3x=\sqrt{4x-3}\\x=\sqrt{4x-3}\end{matrix}\right.\)
\(\Leftrightarrow\left[\begin{matrix}9x^2-4x+3=0\\x^2-4x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[\begin{matrix}x=1\\x=3\end{matrix}\right.\)
3.\(pt\Leftrightarrow\sqrt{3x+8}-\sqrt{3x+5}=\sqrt{5x-4}-\sqrt{5x-7}\)
\(\Leftrightarrow\frac{3x+8-5x+4}{\sqrt{3x+8}+\sqrt{5x+4}}-\frac{3x+5-5x+7}{\sqrt{3x+5}+\sqrt{5x+7}}=0\)
\(\Leftrightarrow\left(12-2x\right)\left(\frac{1}{\sqrt{3x+8}+\sqrt{5x+4}}+\frac{1}{\sqrt{3x+5}+\sqrt{5x+7}}\right)=0\)
\(\Rightarrow x=6\)
a/ đk: \(\left[{}\begin{matrix}x\le\frac{-5-3\sqrt{5}}{10}\\x\ge\frac{-5+3\sqrt{5}}{10}\end{matrix}\right.\)\(\sqrt{x^2+x+1}+\sqrt{3x^2+3x+2}=\sqrt{5x^2+5x-1}\)
\(\Leftrightarrow\sqrt{x^2+x+1}+\sqrt{3\left(x^2+x+1\right)-1}=\sqrt{5\left(x^2+x+1\right)-6}\)
đặt\(x^2+x+1=t\left(t>0\right)\)
\(\sqrt{t}+\sqrt{3t-1}=\sqrt{5t-6}\)
bình phương 2 vế pt trở thành:
\(t+3t-1+2\sqrt{t\left(3t-1\right)}=5t-6\)
\(\Leftrightarrow2\sqrt{3t^2-t}=t-5\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge5\\\left(2\sqrt{3t^2-t}\right)^2=\left(t-5\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge5\\11t^2+6t-25=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge5\\\left[{}\begin{matrix}t=\frac{-3+2\sqrt{71}}{11}\\t=\frac{-3-2\sqrt{71}}{11}\end{matrix}\right.\end{matrix}\right.\)=> không có gtri t nào t/m
vậy pt vô nghiệm
a/ ĐKXĐ: ...
Đặt \(x^2+x+1=a>0\)
\(\sqrt{a}+\sqrt{3a-1}=\sqrt{5a-6}\)
\(\Leftrightarrow4a-1+2\sqrt{3a^2-a}=5a-6\)
\(\Leftrightarrow2\sqrt{3a^2-a}=a-5\) (\(a\ge5\))
\(\Leftrightarrow4\left(3a^2-a\right)=a^2-10a+25\)
\(\Leftrightarrow11a^2+6a-25=0\)
Nghiệm xấu quá, chắc bạn nhầm lẫn đâu đó
b/
Đặt \(x^2+x+1=a>0\)
\(\sqrt{a+3}+\sqrt{a}=\sqrt{2a+7}\)
\(\Leftrightarrow2a+3+2\sqrt{a^2+3a}=2a+7\)
\(\Leftrightarrow\sqrt{a^2+3a}=2\)
\(\Leftrightarrow a^2+3a-4=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-4\left(l\right)\end{matrix}\right.\)
\(\Rightarrow x^2+x+1=1\)
a/ ĐKXĐ: \(-\frac{1}{2}\le x\le4\)
\(\sqrt{4-x}=\sqrt{x+1}+\sqrt{2x+1}\)
\(\Leftrightarrow4-x=3x+2+2\sqrt{2x^2+3x+1}\)
\(\Leftrightarrow1-2x=\sqrt{2x^2+3x+1}\) (\(x\le\frac{1}{2}\))
\(\Leftrightarrow4x^2-4x+1=2x^2+3x+1\)
\(\Leftrightarrow2x^2-7x=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\frac{7}{2}\left(l\right)\end{matrix}\right.\)
Bài này liên hợp cũng được
b/ ĐKXĐ: ...
\(\Leftrightarrow\sqrt{5x+1}^2-\sqrt{5x+1}\left(\sqrt{14x+7}-\sqrt{2x+3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+1=0\Rightarrow x=-\frac{1}{5}\\\sqrt{5x+1}-\sqrt{14x+7}+\sqrt{2x+3}=0\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\sqrt{5x+1}+\sqrt{2x+3}=\sqrt{14x+7}\)
\(\Leftrightarrow7x+4+2\sqrt{10x^2+17x+3}=14x+7\)
\(\Leftrightarrow2\sqrt{10x^2+17x+3}=7x+3\)
\(\Leftrightarrow4\left(10x^2+17x+3\right)=\left(7x+3\right)^2\)
\(\Leftrightarrow...\)
c/ ĐKXĐ: \(x\ge\frac{1}{2}\)
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{2-2x}=a\\\sqrt{2x-1}=b\end{matrix}\right.\) ta được:
\(\left\{{}\begin{matrix}a=1-b\\a^3+b^2=1\end{matrix}\right.\) \(\Rightarrow a^3+\left(1-a\right)^2=1\)
\(\Leftrightarrow a^3+a^2-2a=0\)
\(\Leftrightarrow a\left(a^2+a-2\right)=0\Rightarrow\left[{}\begin{matrix}a=0\\a=1\\a=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2-2x=0\\2-2x=1\\2-2x=-8\end{matrix}\right.\)
d/ ĐKXĐ: \(x\le\frac{5}{4}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{5-4x}=a\\\sqrt[3]{x+7}=b\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=3\\a^2+4b^3=33\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=3-b\\a^2+4b^3=33\end{matrix}\right.\)
\(\Leftrightarrow\left(3-b\right)^2+4b^3=33\)
\(\Leftrightarrow4b^3+b^2-6b-24=0\)
\(\Leftrightarrow\left(b-2\right)\left(4b^2+9b+12\right)=0\)
\(\Rightarrow b=2\Rightarrow\sqrt[3]{x+7}=2\Rightarrow x=1\)
a/ \(\Leftrightarrow\sqrt{x^2+x+3}-\sqrt{x^2+2}+\sqrt{x^2+x+8}-\sqrt{x^2+7}=0\)
\(\Leftrightarrow\frac{x+1}{\sqrt{x^2+x+3}+\sqrt{x^2+2}}+\frac{x+1}{\sqrt{x^2+x+8}+\sqrt{x^2+7}}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{\sqrt{x^2+x+3}+\sqrt{x^2+2}}+\frac{1}{\sqrt{x^2+x+8}+\sqrt{x^2+7}}\right)=0\)
\(\Leftrightarrow x+1=0\) (ngoặc to phía sau luôn dương)
\(\Rightarrow x=-1\)
b/
\(\sqrt{7-x^2+x\sqrt{x+5}}=\sqrt{3-2x-x^2}\) (1)
\(\Rightarrow7-x^2+x\sqrt{x+5}=3-2x-x^2\)
\(\Leftrightarrow x\sqrt{x+5}=-2x-4\)
\(\Rightarrow x^2\left(x+5\right)=4x^2+16x+16\)
\(\Rightarrow x^3+x^2-16\left(x+1\right)=0\)
\(\Rightarrow\left(x+1\right)\left(x^2-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\\x=-2\end{matrix}\right.\)
Do các phép biến đổi ko tương đương nên cần thay nghiệm vào (1) để kiểm tra
c/ ĐKXĐ: \(x\ge\frac{5}{3}\)
\(\Leftrightarrow\sqrt{10x+1}-\sqrt{9x+4}+\sqrt{3x-5}-\sqrt{2x-2}=0\)
\(\Leftrightarrow\frac{x-3}{\sqrt{10x+1}+\sqrt{9x+4}}+\frac{x-3}{\sqrt{3x-5}+\sqrt{2x-2}}=0\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{1}{\sqrt{10x+1}+\sqrt{9x+4}}+\frac{1}{\sqrt{3x-5}+\sqrt{2x-2}}\right)=0\)
\(\Leftrightarrow x-3=0\) (ngoặc phía sau luôn dương)
d/ Đề bài là \(2\sqrt{2x+3}\) hay \(2\sqrt{2x-3}\) bạn?
e/ ĐKXĐ: \(x\ge-3\)
\(\Leftrightarrow\sqrt{x+3+2\sqrt{x+3}+1}=x+4\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x+3}+1\right)^2}=x+4\)
\(\Leftrightarrow\sqrt{x+3}+1=x+4\)
\(\Leftrightarrow x+3-\sqrt{x+3}=0\)
\(\Leftrightarrow\sqrt{x+3}\left(\sqrt{x+3}-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x+3=1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\)
28. \(x^2+\frac{9x^2}{\left(x-3\right)^2}=40\) DK: \(x\ne3\)
PT\(\Leftrightarrow\left(x+\frac{3x}{x-3}\right)^2-6\frac{x^2}{x-3}-40=0\)\(\Leftrightarrow\frac{x^4}{\left(x-3\right)^2}-6\frac{x^2}{x-3}-40=0\)
Dat \(\frac{x^2}{x-3}=a\). PTTT \(a^2-6a-40=0\)\(\Leftrightarrow\left(a-10\right)\left(a+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=10\\a=-4\end{matrix}\right.\)
giai tiep
14. \(\frac{1}{\sqrt{x}+1}+\frac{1}{\sqrt{x}-1}=1\) DK: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
PT\(\Leftrightarrow\frac{\sqrt{x}-1+\sqrt{x}+1}{x-1}=1\Leftrightarrow2\sqrt{x}=x-1\)\(\Leftrightarrow x-2\sqrt{x}+1=2\Leftrightarrow\left(\sqrt{x}-1\right)^2=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3+2\sqrt{2}\\x=3-2\sqrt{2}\end{matrix}\right.\)
a) ĐK : \(x\ge\frac{2}{3}\)\(\sqrt{3x-2}-\sqrt{x+7}=1\Leftrightarrow3x-2-2\sqrt{\left(3x-2\right)\left(x+7\right)}+x+7=1\)
\(\Leftrightarrow4x+5-1=2\sqrt{3x^2+19x-14}\Leftrightarrow2x+2=\sqrt{3x^2+19x-14}\)
\(\Leftrightarrow4x^2+8x+4=3x^2+19x-14\)
\(\Leftrightarrow x^2-11x+18=0\Leftrightarrow\left[{}\begin{matrix}x=9\left(tm\right)\\x=2\left(tm\right)\end{matrix}\right.\)
b) ĐK \(x\ge-\frac{1}{5}\)\(\sqrt{14x+7}-\sqrt{2x+3}=\sqrt{5x+1}\Leftrightarrow14x+7+2x+3-5x-1-2\sqrt{28x^2+42x+14x+21}=0\)
\(\Leftrightarrow11x+9=2\sqrt{28x^2+56x+21}\Leftrightarrow121x^2+81+198x=112x^2+224x+84\)
\(\Leftrightarrow9x^2-26x-3=0\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-\frac{1}{9}\left(loai\right)\end{matrix}\right.\)
c) \(\sqrt{x^2+2x+6}-\sqrt{x^2+x+2}=1\)
\(\Leftrightarrow x^2+2x+6=x^2+x+2+1+2\sqrt{x^2+x+2}\)
\(\Leftrightarrow x+3=2\sqrt{x^2+x+2}\)
\(\Leftrightarrow x^2+6x+9=4x^2+4x+8\)
\(\Leftrightarrow3x^2-2x-1=0\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=-\frac{1}{3}\left(tm\right)\end{matrix}\right.\)