giải pt
\(\dfrac{140}{x}+5=\dfrac{\left(140+10\right)}{x-1}\)
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NV
Nguyễn Việt Lâm
Giáo viên
21 tháng 2 2021
\(\Leftrightarrow10\left(x^2+\dfrac{1}{x^2}+2\right)+5\left(x^2+\dfrac{1}{x^2}\right)^2-5\left(x^2+\dfrac{1}{x^2}\right)\left(x^2+\dfrac{1}{x^2}+2\right)=\left(x-5\right)^2-5\)
\(\Leftrightarrow10\left(x^2+\dfrac{1}{x^2}\right)+20+5\left(x^2+\dfrac{1}{x^2}\right)^2-5\left(x^2+\dfrac{1}{x^2}\right)^2-10\left(x^2+\dfrac{1}{x^2}\right)=\left(x-5\right)^2-5\)
\(\Leftrightarrow\left(x-5\right)^2=25\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=10\end{matrix}\right.\)
17 tháng 2 2022
a: \(\Leftrightarrow x^2+x-6+2x-6=10x-20+50\)
\(\Leftrightarrow x^2+3x-12-10x-30=0\)
\(\Leftrightarrow x^2-7x-42=0\)
\(\text{Δ}=\left(-7\right)^2-4\cdot1\cdot\left(-42\right)=217>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{7-\sqrt{217}}{2}\\x_2=\dfrac{7+\sqrt{217}}{2}\end{matrix}\right.\)
b: \(\Leftrightarrow x^2-3x+5=-x^2+4\)
\(\Leftrightarrow2x^2-3x+1=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x-1\right)=0\)
hay \(x\in\left\{\dfrac{1}{2};1\right\}\)
HN
2
20 tháng 5 2022
\(\Leftrightarrow x-1-4=5\left(x-5\right)\)
=>x-5=5(x-5)
=>x-5-5x+25=0
=>-4x+20=0
hay x=5(loại)
1 tháng 1 2022
\(ĐK:x\ne2;x\ne-3\\ PT\Leftrightarrow\left(x-2\right)\left(x+3\right)+2\left(x+3\right)=10\left(x-2\right)+50\\ \Leftrightarrow x^2+x-6+2x+6=10x-20+50\\ \Leftrightarrow x^2-13x-30=0\\ \Leftrightarrow x^2-15x+2x-30=0\\ \Leftrightarrow\left(x-15\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=15\\x=-2\end{matrix}\right.\left(tm\right)\)
16 tháng 4 2017
Câu 1:
Ta có: \(\dfrac{x-4}{y-3}=\dfrac{4}{3}\)
=> \(3.\left(x-4\right)=4.\left(y-3\right)\)
=>\(3x-12=4y-12\)
=>\(3x=4y\) (1)
Ta có: \(x-y=5\)
=> \(y=y+5\) Thay vào (1) ta có:
\(3.\left(y+5\right)=4.\)y
=>\(3y+15=4y\)
=> \(15=4y-3y\)
=> 15 = y
=> y =15
ta có: x = y +5
=> x = 15 +5
=> x =20
16 tháng 4 2017
Câu 2:
\(B=\dfrac{10}{56}+\dfrac{10}{140}+\dfrac{10}{260}+...+\dfrac{10}{1400}\)
\(B=\dfrac{5}{28}+\dfrac{6}{70}+\dfrac{5}{130}+...+\dfrac{5}{700}\)
\(B=\dfrac{5}{4.7}+\dfrac{5}{7.10}+\dfrac{5}{10.13}+...+\dfrac{5}{25.28}\)
\(B=5,\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+...+\dfrac{1}{25}-\dfrac{1}{28}\right)\)
\(3B=5.\left(\dfrac{3}{4.7}+\dfrac{3}{7.10}+\dfrac{3}{10.13}+...+\dfrac{3}{25.28}\right)\)
\(3B=5.\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+...+\dfrac{1}{25}-\dfrac{1}{28}\right)\)
\(3B=5.\left(\dfrac{1}{4}-\dfrac{1}{28}\right)\)
\(3B=5.\dfrac{3}{14}\)
\(B=\dfrac{15}{14}:3=\dfrac{5}{14}\)
Câu 3:
38 - (|x+10|+13) = \(\left(-6\right)^{20}:\left(9^9.4^{10}\right)\)
=> \(38-\left(\left|x+10\right|+13\right)=\left(2.3\right)_{ }^{20}:\)\(\left[\left(3^2\right)^9.\left(2^2\right)^4\right]\)
=>\(38-\left(\left|x+10\right|+13\right)=2^{20}.3^{20}:\left(3^{18}.2^{20}\right)\)
=> \(38-\left(\left|x+10\right|+13\right)=\dfrac{3^{20}.2^{20}}{3^{18}.2^{20}}\)
=> \(38-\left(\left|x+10\right|+13\right)=9\)
=> |x +10| + 13 = 38 -9
=> |x+10| +13 = 29
=> |x+10| = 29 -13
=> |x+10| = 16
\(\Rightarrow\left[{}\begin{matrix}x+10=16\\x+10=-16\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=6\\x=-26\end{matrix}\right.\)
18 tháng 6 2022
\(\Leftrightarrow\left(\dfrac{x-1}{99}-1\right)+\left(\dfrac{x-99}{1}-1\right)+\left(\dfrac{x-3}{97}-1\right)+\left(\dfrac{x-7}{93}-1\right)+\left(\dfrac{x-5}{95}-1\right)+\left(\dfrac{x-95}{5}-1\right)=0\)=>x-100=0
hay x=100
5 tháng 3 2021
a) Sửa đề: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)
ĐKXĐ: \(x\notin\left\{3;\dfrac{1}{5}\right\}\)
Ta có: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)
\(\Leftrightarrow\dfrac{3\left(3-x\right)}{\left(5x-1\right)\left(3-x\right)}+\dfrac{2\left(5x-1\right)}{\left(3-x\right)\left(5x-1\right)}=\dfrac{4}{\left(5x-1\right)\left(3-x\right)}\)
Suy ra: \(9-3x+10x-2=4\)
\(\Leftrightarrow7x+7=4\)
\(\Leftrightarrow7x=-3\)
hay \(x=-\dfrac{3}{7}\)
Vậy: \(S=\left\{-\dfrac{3}{7}\right\}\)
\(\dfrac{140}{x}+5=\dfrac{\left(140+10\right)}{x-1}\left(x\ne0,x\ne1\right)\)
\(\Leftrightarrow\dfrac{140+5x}{x}=\dfrac{150}{x-1}\)
\(\Leftrightarrow\left(x-1\right)\cdot\left(140+5x\right)=150x\)
\(\Leftrightarrow140x+5x^2-140-5x-150x=0\)
\(\Leftrightarrow5x^2-15x-140=0\)
\(\Leftrightarrow x^2-3x-28=0\)
\(\Leftrightarrow\left(x-7\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\left(N\right)\\x=-4\left(N\right)\end{matrix}\right.\)
\(S=\left\{7,-4\right\}\)
ĐK: `x \ne 0 ; x \ne -1`
`140/x+5=150/(x-1)`
`<=>(140+5x)/x=150/(x-1)`
`<=>(140x+5x)(x-1)=150x`
`<=>5x^2+135x-140=150x`
`<=>5x^2-15x-140=0`
`<=>` \(\left[{}\begin{matrix}x=7\\x=-4\end{matrix}\right.\)
Vậy...