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10 tháng 7 2021
Ta có: \(\left(\dfrac{2}{\sqrt{x}-2}+\dfrac{3}{2\sqrt{x}+1}-\dfrac{5\sqrt{x}-7}{2x-3\sqrt{x}-2}\right):\dfrac{2\sqrt{x}+3}{5x-10\sqrt{x}}\)
\(=\dfrac{4\sqrt{x}+2+3\sqrt{x}-6-5\sqrt{x}+7}{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)
\(=\dfrac{2\sqrt{x}+3}{2\sqrt{x}+1}\cdot\dfrac{5\sqrt{x}}{2\sqrt{x}+3}\)
\(=\dfrac{5\sqrt{x}}{2\sqrt{x}+1}\)
25 tháng 6 2019
TL:
1đk:x<1
.\(1+3x-1=9x^2\)
\(3x=9x^2\)
x=3x\(^2\)
=>x=0(ktm) hoặc x= \(\frac{1}{3}\left(tm\right)\)
vậy x=\(\frac{1}{3}\)
hc tốt:)
CT
12 tháng 2 2020
ĐKXĐ: x>4
A= \(\left(\frac{2}{\sqrt{x}-2}+\frac{3}{2\sqrt{x}+1}-\frac{5\sqrt{x}-7}{2x-3\sqrt{x}-2}\right):\frac{2\sqrt{x}+3}{5x-10\sqrt{x}}\)
A= \(\left[\frac{2\left(2\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}+\frac{3\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}-\frac{5\sqrt{x}-7}{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}\right]\cdot\frac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)
A= \(\left[\frac{4\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}+\frac{3\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}-\frac{5\sqrt{x}-7}{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}\right]\)\(\cdot\frac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)
A= \(\frac{2\sqrt{x}+3}{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}\cdot\frac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)\(=\frac{5\sqrt{x}}{2\sqrt{x}+1}\)\(=2,5-\frac{2,5}{2\sqrt{x}+1}\)
...
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 4 2019
a/
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{7-x}=a\\\sqrt[3]{x-5}=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a^3+b^3=3\\a^3-b^3=2\left(6-x\right)\end{matrix}\right.\) với \(a+b\ne0\)
Ta có hệ:
\(\left\{{}\begin{matrix}a^3+b^3=2\\\frac{a-b}{a+b}=\frac{a^3-b^3}{2}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a^3+b^3=2\\\frac{a-b}{a+b}=\frac{\left(a-b\right)\left(a^2+ab+b^2\right)}{2}\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}a^3+b^3=2\\a-b=0\end{matrix}\right.\) \(\Rightarrow a=b=1\Rightarrow\left\{{}\begin{matrix}\sqrt[3]{7-x}=1\\\sqrt[3]{x-5}=1\end{matrix}\right.\) \(\Rightarrow x=6\)
TH2: \(\left\{{}\begin{matrix}a^3+b^3=2\\\frac{1}{a+b}=\frac{a^2+ab+b^2}{2}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a^3+b^3=2\\\frac{1}{a+b}=\frac{a^2+ab+b^2}{a^3+b^3}=\frac{a^2+ab+b^2}{\left(a+b\right)\left(a^2-ab+b^2\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a^3+b^3=2\\\frac{a^2+ab+b^2}{a^2-ab+b^2}=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a^3+b^3=2\\ab=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a=0\\b^3=2\end{matrix}\right.\\\left\{{}\begin{matrix}b=0\\a^3=2\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=7\\x=5\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 4 2019
b/
Lập phương 2 vế:
\(\left(\sqrt[3]{x+1}+\sqrt[3]{x-1}\right)^3=5x\)
\(\Leftrightarrow x+1+x-1+3\sqrt[3]{\left(x^2-1\right)}\left(\sqrt[3]{x+1}+\sqrt[3]{x-1}\right)=5x\)
\(\Leftrightarrow2x+3\sqrt[3]{x^2-1}\left(\sqrt[3]{5x}\right)=5x\)
\(\Leftrightarrow x=\sqrt[3]{5x\left(x^2-1\right)}\)
\(\Leftrightarrow x^3=5x\left(x^2-1\right)\)
\(\Leftrightarrow x\left(5\left(x^2-1\right)-x^2\right)=0\)
\(\Leftrightarrow x\left(4x^2-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{\sqrt{5}}{2}\\x=-\frac{\sqrt{5}}{2}\end{matrix}\right.\)
23 tháng 10 2020
1. ĐKXĐ: \(x\ge3\)
\(2\sqrt{9x-27}-\frac{1}{5}\sqrt{25x-75}-\frac{1}{7}\sqrt{49x-147}=20\)
⇔ \(6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20\)
⇔ \(4\sqrt{x-3}=20\)
⇔ \(\sqrt{x-3}=5\)
⇔ \(x-3=25\)
⇔ \(x=28\left(TMĐKXĐ\right)\)
Vậy....
23 tháng 10 2020
2. ĐKXĐ: \(x\ge0\)
\(\frac{3}{2}\sqrt{5x}+\sqrt{5x}-7=\frac{1}{2}\sqrt{5x}\)
⇔ \(\frac{3}{2}\sqrt{5x}+\sqrt{5x}-\frac{1}{2}\sqrt{5x}=7\)
⇔ \(2\sqrt{5x}=7\)
⇔ \(\sqrt{5x}=\frac{7}{2}\)
⇔ \(5x=\frac{49}{4}\)
⇔ \(x=\frac{49}{20}\left(TMĐKXĐ\right)\)
Vậy...
6 tháng 10 2020
1) đk: \(x\ge1\)
Ta có: \(\sqrt{x-1}-\sqrt{2x\left(x-1\right)}=0\)
\(\Leftrightarrow\sqrt{x-1}=\sqrt{2x\left(x-1\right)}\)
\(\Leftrightarrow x-1=2x^2-2x\)
\(\Leftrightarrow2x^2-3x+1=0\)
\(\Leftrightarrow\left(2x^2-2x\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\left(ktm\right)\\x=1\left(tm\right)\end{cases}}\)
Vậy x = 1
2) đk: \(x\ge\frac{1}{2}\)
Ta có: \(\sqrt{5x^2}=2x-1\)
\(\Leftrightarrow5x^2=\left(2x-1\right)^2\)
\(\Leftrightarrow5x^2=4x^2-4x+1\)
\(\Leftrightarrow x^2+4x-1=0\)
\(\Leftrightarrow\left(x+2\right)^2-5=0\)
\(\Leftrightarrow\left(x+2-\sqrt{5}\right)\left(x+2+\sqrt{5}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-2+\sqrt{5}\left(ktm\right)\\x=-2-\sqrt{5}\left(ktm\right)\end{cases}}\)
=> PT vô nghiệm
3) đk: \(x\ge-1\)
Ta có: \(\sqrt{x+1}+\sqrt{9x+9}=4\)
\(\Leftrightarrow\sqrt{x+1}+3\sqrt{x+1}=4\)
\(\Leftrightarrow4\sqrt{x+1}=4\)
\(\Leftrightarrow x+1=1\)
\(\Rightarrow x=0\)
6 tháng 10 2020
4) đk: \(x\ge2\)
Ta có: \(\sqrt{x-2}-\sqrt{x\left(x-2\right)}=0\)
\(\Leftrightarrow\sqrt{x-2}=\sqrt{x\left(x-2\right)}\)
\(\Leftrightarrow x-2=x\left(x-2\right)\)
\(\Leftrightarrow x\left(x-2\right)-\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\left(ktm\right)\\x=2\left(tm\right)\end{cases}}\)
Vậy x = 2
6) đk: \(x\ge-\frac{7}{5}\)
Ta có: \(\frac{\sqrt{2x-3}}{\sqrt{x-1}}=2\)
\(\Leftrightarrow\frac{2x-3}{x-1}=2\)
\(\Leftrightarrow2x-3=2x-2\)
\(\Leftrightarrow0x=1\) vô lý
=> PT vô nghiệm
x=0 thế củng hỏi
x>=-3>-7/5