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3 tháng 9 2022
b: \(\Leftrightarrow\dfrac{\left(x+2\right)^2}{\left(x-1\right)\left(x+2\right)}=\dfrac{-4x^2+11x-2}{\left(x+2\right)\left(x-1\right)}\)
\(\Leftrightarrow x^2+4x+4+4x^2-11x+2=0\)
\(\Leftrightarrow5x^2-7x+6=0\)
hay \(x\in\varnothing\)
c: \(\Leftrightarrow\left(3x^2+2\right)^2-5x\left(3x^2+2\right)=0\)
=>3x^2-5x+2=0
=>3x^2-3x-2x+2=0
=>(x-1)(3x-2)=0
=>x=2/3 hoặc x=1
21 tháng 4 2017
\(\dfrac{1}{x}+\dfrac{1}{x+50}=\dfrac{1}{60}\left(x\ne0;x\ne-5\right)\)
\(pt\Leftrightarrow\dfrac{x+50}{x\left(x+50\right)}+\dfrac{x}{x\left(x+50\right)}=\dfrac{1}{60}\)
\(\Leftrightarrow\dfrac{2x+50}{x\left(x+50\right)}=\dfrac{1}{60}\Leftrightarrow x\left(x+50\right)=60\left(2x+50\right)\)
\(\Leftrightarrow x^2+50x=120x+3000\)
\(\Leftrightarrow x^2-70x-3000=0\)
\(\Leftrightarrow x^2-100x+30x-3000=0\)
\(\Leftrightarrow x\left(x-100\right)+30\left(x-100\right)=0\)
\(\Leftrightarrow\left(x+30\right)\left(x-100\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+30=0\\x-100=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-30\\x=100\end{matrix}\right.\)
CV
1
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
3 tháng 5 2023
\(\dfrac{100}{x}-\dfrac{100}{x+10}=\dfrac{30}{60}=0,5\left(ĐKXĐ:x\ne0;x\ne-10\right)\\ \Leftrightarrow\dfrac{100\left(x+10\right)-100x}{x\left(x+10\right)}=\dfrac{0,5x\left(x+10\right)}{x\left(x+10\right)}\\ \Leftrightarrow100x-100x+1000=0,5x^2+5x\\ \Leftrightarrow0,5x^2+5x-1000=0\\ \Leftrightarrow0,5x^2-20x+25x-1000=0\\ \Leftrightarrow0,5x.\left(x-40\right)+25.\left(x-40\right)=0\\ \Leftrightarrow\left(0,5x+25\right)\left(x-40\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}0,5x+25=0\\x-40=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-50\\x=40\end{matrix}\right.\\ Vậy:S=\left\{-50;40\right\}\)
18 tháng 8 2021
Bài 2:
a: Ta có: \(\sqrt{8-2\sqrt{15}}\cdot\left(\sqrt{60}+6\right):2\sqrt{3}\)
\(=\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{12}\left(\sqrt{5}+\sqrt{3}\right):2\sqrt{3}\)
\(=2\sqrt{12}:2\sqrt{3}\)
=2
b: Ta có: \(\sqrt{5-\sqrt{21}}-\sqrt{\dfrac{7}{2}}\)
\(=\dfrac{\sqrt{10-2\sqrt{21}}-\sqrt{7}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{7}-\sqrt{3}-\sqrt{7}}{\sqrt{2}}\)
\(=-\dfrac{\sqrt{6}}{2}\)
21 tháng 2 2023
a: \(\Leftrightarrow x^2+x+1-3x^2=2x\left(x-1\right)\)
=>-2x^2+x+1-2x^2+2x=0
=>-4x^2+3x+1=0
=>4x^2-3x-1=0
=>4x^2-4x+x-1=0
=>(x-1)(4x+1)=0
=>x=1(loại) hoặc x=-1/4(nhận)
b: \(\Leftrightarrow\dfrac{440}{x-2}-\dfrac{440}{x}=1\)
=>x(x-2)=440x-440x+880
=>x^2-2x-880=0
=>\(x=1\pm\sqrt{881}\)
c: \(\Leftrightarrow\dfrac{x+5+x}{x\left(x+5\right)}=\dfrac{1}{6}\)
=>x^2+5x=6(2x+5)
=>x^2+5x-12x-30=0
=>x^2-7x-30=0
=>(x-10)(x+3)=0
=>x=10 hoặc x=-3
d: =>(x-1)(x+1)-x=2x-1
=>x^2-1-x=2x-1
=>x^2-x-2x=0
=>x(x-3)=0
=>x=0(loại) hoặc x=3(nhận)
AH
Akai Haruma
Giáo viên
4 tháng 11 2017
Lời giải:
PT (1)\(\rightarrow x_1+x_2=\frac{60.3}{4}=45\)
\(\Rightarrow x_2=45-x_1\)
Thay vào pt (2)
\(\frac{60}{x_2}-\frac{60}{x_1}=2\)
\(\Leftrightarrow \frac{60}{45-x_1}-\frac{60}{x_1}=2\)
\(\Leftrightarrow \frac{1}{45-x_1}-\frac{1}{x_1}=\frac{1}{30}\Leftrightarrow \frac{x_1-(45-x_1)}{x_1(45-x_1)}=\frac{1}{30}\)
\(\Leftrightarrow 30(2x_1-45)=x_1(45-x_1)\)
\(\Leftrightarrow x_1^2+15x_1-1350=0\)
\(\Rightarrow\left[{}\begin{matrix}x_1=30\rightarrow x_2=15\\x_1=-45\rightarrow x_2=90\end{matrix}\right.\)
(đều thỏa mãn)
Vậy \((x_1,x_2)=(30;15);(-45;90)\)
1 tháng 1 2023
a: \(=\sqrt{\dfrac{1}{10}}+\sqrt{\dfrac{1}{60}}-\dfrac{2\sqrt{15}}{15}\)
\(=\dfrac{\sqrt{10}}{10}-\dfrac{2\sqrt{15}}{15}+\dfrac{\sqrt{15}}{30}\)
\(=\dfrac{3\sqrt{10}-3\sqrt{15}}{30}=\dfrac{\sqrt{10}-\sqrt{15}}{10}\)
b: \(=\dfrac{\left(\sqrt{5}+\dfrac{1}{2}\cdot2\sqrt{5}-\dfrac{5}{4}\cdot\dfrac{2}{\sqrt{5}}+\sqrt{5}\right)}{2\sqrt{5}}\)
\(=\dfrac{\left(\sqrt{5}+\sqrt{5}-\dfrac{1}{2}\sqrt{5}+\sqrt{5}\right)}{2\sqrt{5}}\)
\(=\dfrac{5}{2}:2=\dfrac{5}{4}\)
9 tháng 8 2023
a: \(M=1:\left(\dfrac{1}{\sqrt{x}+2}-\dfrac{3x}{2\left(x-4\right)}+\dfrac{1}{2\left(\sqrt{x}-2\right)}\right)\cdot\dfrac{4-2\sqrt{x}}{1}\)
\(=1:\left(\dfrac{2\sqrt{x}-4-3x+\sqrt{x}+2}{2\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right)\cdot\dfrac{-2\left(\sqrt{x}-2\right)}{1}\)
\(=\dfrac{2\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)\cdot\left(-2\right)\cdot\left(\sqrt{x}-2\right)}{-3x+3\sqrt{x}-2}\)
\(=\dfrac{-4\left(\sqrt{x}-2\right)^2\left(\sqrt{x}+2\right)}{-3x+3\sqrt{x}-2}\)
b: M=20
=>\(-4\left(x-4\right)\left(\sqrt{x}-2\right)=-60x+60\sqrt{x}-40\)
=>\(x\sqrt{x}-2x-4\sqrt{x}+8=-15x+15\sqrt{x}-10\)
=>\(x\sqrt{x}+13x-19\sqrt{x}+18=0\)
=>\(x\in\varnothing\)
\(\dfrac{60}{x}-\dfrac{60}{x+2}=\dfrac{1}{20}\left(đk:x\ne0,x\ne-2\right)\)
\(\Leftrightarrow\dfrac{60x+120-60x}{x\left(x+2\right)}=\dfrac{1}{20}\)
\(\Leftrightarrow\dfrac{120}{x^2+2x}=\dfrac{1}{20}\Leftrightarrow x^2+2x=2400\)
\(\Leftrightarrow\left(x+1\right)^2=2401\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=49\\x+1=-49\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=48\\x=-50\end{matrix}\right.\)(thỏa đk)
Ta có: \(\dfrac{60}{x}-\dfrac{60}{x+2}=\dfrac{1}{20}\)
\(\Leftrightarrow x\left(x+2\right)=1200x+2400-1200x\)
\(\Leftrightarrow x^2+2x-2400=0\)
\(\Delta=2^2-4\cdot1\cdot\left(-2400\right)=9604\)
Vì \(\Delta>0\) nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{-2-98}{2}=-50\left(nhận\right)\\x_2=\dfrac{-2+98}{2}=48\left(nhận\right)\end{matrix}\right.\)