giải pt : \(\sqrt[11]{x-4}+\sqrt[11]{x-5}+\sqrt[11]{2x-9}=2\)
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a, ĐK: \(x\ge11\)
\(\sqrt{x+\sqrt{x-11}}+\sqrt{x-\sqrt{x-11}}=4\)
\(\Leftrightarrow x+\sqrt{x-11}+x-\sqrt{x-11}+2\sqrt{x^2-x+11}=16\)
\(\Leftrightarrow2x+2\sqrt{x^2-x+11}=16\)
\(\Leftrightarrow x+\sqrt{x^2-x+11}=8\)
Ta thấy \(x+\sqrt{x^2-x+11}>11>\text{}8\)
\(\Rightarrow\) phương trình vô nghiệm.
\(a,\sqrt{x+\sqrt{x-11}}+\sqrt{x-\sqrt{x-11}}=4\left(x\ge11\right)\\ \Leftrightarrow x+\sqrt{x-11}+x-\sqrt{x-11}+2\sqrt{\left(x+\sqrt{x-11}\right)\left(x-\sqrt{x-11}\right)}=16\\ \Leftrightarrow2x+2\sqrt{x^2-x+11}=16\\ \Leftrightarrow x+\sqrt{x^2-x+11}=8\\ \Leftrightarrow\sqrt{x^2-x+11}=8-x\\ \Leftrightarrow x^2-x+11=x^2-16x+64\\ \Leftrightarrow15x=53\\ \Leftrightarrow x=\dfrac{53}{15}\left(ktm\right)\)
\(b,\sqrt{x+2+3\sqrt{2x-5}}+\sqrt{x-2-\sqrt{2x-5}}=2\sqrt{2}\left(x\ge\dfrac{5}{2}\right)\\ \Leftrightarrow\sqrt{2x-5+6\sqrt{2x-5}+9}+\sqrt{2x-5-2\sqrt{2x-5}+1}=4\\ \Leftrightarrow\sqrt{\left(\sqrt{2x-5}+3\right)^2}+\sqrt{\left(\sqrt{2x-5}-1\right)^2}=4\\ \Leftrightarrow\sqrt{2x-5}+3+\left|\sqrt{2x-5}-1\right|=4\\ \Leftrightarrow\left|\sqrt{2x-5}-1\right|=1-\sqrt{2x-5}\\ \Leftrightarrow\sqrt{2x-5}-1\le0\\ \Leftrightarrow\sqrt{2x-5}\le1\\ \Leftrightarrow2x-5\le1\Leftrightarrow x\le\dfrac{5}{2}\\ \Leftrightarrow x=\dfrac{5}{2}\)
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2(sin2xcos\(\frac{9\pi}{4}\) + sin\(\frac{9\pi}{4}\)cosx) + 7\(\sqrt{2}\)sinx + \(\sqrt{2}\)( sinx cos\(\frac{11\pi}{2}\)+sin\(\frac{11\pi}{2}\)cosx ) =4\(\sqrt{2}\)
\(\sqrt{2}\)sin2x + \(\sqrt{2}\)cosx +7\(\sqrt{2}\)sinx -\(\sqrt{2}\)cosx =4\(\sqrt{2}\)
2\(\sqrt{2}\)sinxcosx+7\(\sqrt{2}\)sinx - 4\(\sqrt{2}\) =0
PHẦN CÒN LẠI C TỰ LM NỐT NHÉ
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\(\left(x-1\right)+4.\left(\sqrt{x+3}-2\right)+2.\left(\sqrt{3-2x}-1\right)=0\)
\(x-1+\dfrac{4.\left(x+3-4\right)}{\sqrt{x+3}+2}+\dfrac{2.\left(3-2x-1\right)}{\sqrt{3-2x}+1}=0\)
=> x-1+\(\dfrac{4.\left(x-1\right)}{\sqrt{x+3}+2}+\dfrac{4.\left(1-x\right)}{\sqrt{3-2x}+1}=0\)
=> (x-1).\(\left(\dfrac{4}{\sqrt{x+3}+2}+\dfrac{4}{\sqrt{3-2x}+1}\right)=0\)
=> x=1 (do \(\dfrac{4}{\sqrt{x+3}+2}+\dfrac{4}{\sqrt{3-2x}+1}>0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(x^2+8=3\sqrt{x^3+8}\)
\(\left(x^2+8\right)^2=\left(3\sqrt{x^2+8}\right)^2\)
\(x^4+16x^2+64=9x^2+72\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)
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ta đặt: \(\sqrt[3]{x+5}=u\)
\(\sqrt[3]{x+6}=v\)
ta có \(u^3+v^3=2x+11\)
=> \(u+v=\sqrt[3]{u^3+v^3}\)
=>\(\left(u+v\right)^3=u^3+v^3+3uv\left(u+v\right)=u^3+v^3\)
=> \(3uv\left(u+v\right)=3uv\sqrt[3]{u^3+v^3}=0\)
<=> \(3\sqrt[3]{x+5}\sqrt[3]{x+6}\sqrt[3]{2x+11}=0\)
<=> x=-5 hoặc x=-6 hoặc x=-11/2
vậy pt có 3 nghiệm ....
![](https://rs.olm.vn/images/avt/0.png?1311)
Quên mất mình đánh nhầm.
ĐKXĐ: \(x\ge-\frac{1}{2}\).
PT đã cho tương đương với:
\(\left(\sqrt{2x+1}-3\right)-\left(\sqrt[3]{x+4}-2\right)=2x^2-5x-12\)
\(\Leftrightarrow\frac{2\left(x-4\right)}{\sqrt{2x+1}+3}-\frac{x-4}{\left(\sqrt[3]{x+4}\right)^2+2\sqrt[3]{x+4}+4}=\left(x-4\right)\left(2x+3\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\Leftrightarrow x=4\\\frac{2}{\sqrt{2x+1}+3}-\frac{1}{\left(\sqrt[3]{x+4}\right)^2+2\sqrt[3]{x+4}+4}=2x+3\left(1\right)\end{matrix}\right.\).
Với \(x\ge-\frac{1}{2}\) ta có: \(VT_{\left(1\right)}\le\frac{2}{3};VP\ge2\).
Do đó (1) vô nghiệm.
Vậy phương trình có nghiệm duy nhất: x = 4.
ĐKXĐ: \(x\ge-\frac{1}{2}\).
PT đã cho tương đương với:
\(\left(\sqrt{2x+1}-3\right)-\left(\sqrt[3]{x+4}-2\right)=2x^2-5x-12\)
\(\Leftrightarrow\frac{2\left(x-4\right)}{\sqrt{2x+1}+3}-\frac{x-4}{\left(\sqrt[3]{x+4}\right)^2+2\sqrt[3]{x+4}+4}=\left(x-4\right)\left(2x+3\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\\frac{1}{\sqrt{2x+1}+3}-\frac{1}{\left(\sqrt[3]{x+4}\right)^2+2\sqrt[3]{x+4}+4}=2x+3\left(1\right)\end{matrix}\right.\).
Với \(x\ge-\frac{1}{2}\) ta có: \(VT_{\left(1\right)}\le\frac{1}{3};VP_{\left(1\right)}\ge2\).
Do đó (1) vô nghiệm.
Vậy x = 4 là nghiệm duy nhất của phương trình.
![](https://rs.olm.vn/images/avt/0.png?1311)
Lập phương hai vế : \(\left(\sqrt[3]{x+5}+\sqrt[3]{x+6}\right)^3=\left(\sqrt[3]{2x+11}\right)^3\)
\(\Leftrightarrow2x+11+3.\sqrt[3]{x+5}.\sqrt[3]{x+6}\left(\sqrt[3]{x+5}+\sqrt[3]{x+6}\right)=2x+11\)
\(\Leftrightarrow\sqrt[3]{x+5}.\sqrt[3]{x+6}\left(\sqrt[3]{x+6}+\sqrt[3]{x+5}\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}\sqrt[3]{x+5}=0\\\sqrt[3]{x+6}=0\\\sqrt[3]{x+5}+\sqrt[3]{x+6}=0\end{array}\right.\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=-5\\x=-6\\x=-\frac{11}{2}\end{array}\right.\)