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10 tháng 5 2023
a:=>3x=15
=>x=5
b: =>8-11x<52
=>-11x<44
=>x>-4
c: \(VT=\left(\dfrac{x^2-\left(x-6\right)^2}{x\left(x+6\right)\left(x-6\right)}\right)\cdot\dfrac{x\left(x+6\right)}{2x-6}+\dfrac{x}{6-x}\)
\(=\dfrac{12x-36}{2x-6}\cdot\dfrac{1}{x-6}-\dfrac{x}{x-6}=\dfrac{6}{x-6}-\dfrac{x}{x-6}=-1\)
23 tháng 2 2022
\(\Leftrightarrow36\left(x+6\right)+36\left(x-6\right)=\dfrac{9}{2}\left(x^2-36\right)\)
\(\Leftrightarrow x^2\cdot\dfrac{9}{2}-162-72x=0\)
\(\Leftrightarrow9x^2-144x-324=0\)
\(\Leftrightarrow x^2-16x-36=0\)
=>(x-18)(x+2)=0
=>x=18 hoặc x=-2
23 tháng 2 2022
ĐKXĐ:\(x\ne\pm6\)
\(\dfrac{36}{x-6}+\dfrac{36}{x+6}=4,5\\ \Leftrightarrow36\left(\dfrac{1}{x-6}+\dfrac{1}{x+6}\right)=4,5\\ \Leftrightarrow\dfrac{x+6}{\left(x-6\right)\left(x+6\right)}+\dfrac{x-6}{\left(x-6\right)\left(x+6\right)}=\dfrac{1}{8}\\ \Leftrightarrow\dfrac{x+6+x-6}{x^2-36}=\dfrac{1}{8}\\ \Leftrightarrow\dfrac{2x}{x^2-36}=\dfrac{1}{8}\\ \Leftrightarrow x^2-36=16x\\ \Leftrightarrow x^2-16x-36=0\\ \Leftrightarrow\left(x^2+2x\right)-\left(18x+36\right)=0\\ \Leftrightarrow x\left(x+2\right)-18\left(x+2\right)=0\\ \Leftrightarrow\left(x+2\right)\left(x-18\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\left(tm\right)\\x=18\left(tm\right)\end{matrix}\right.\)
PT
1
30 tháng 4 2021
\(\dfrac{36}{x+6}+\dfrac{36}{x-6}=4,5\)
\(\Leftrightarrow36\left(x-6\right)+36\left(x+6\right)=4,5\left(x^2-36\right)\)
\(\Leftrightarrow36x-216+36x+216=4,5x^2-162\)
\(\Leftrightarrow-4,5x^2+72x+162=0\)
\(\Leftrightarrow\left(x-18\right)\left(-4,5x-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=18\\x=-2\end{matrix}\right.\)
1 tháng 5 2021
bạn làm rõ hơn ở chỗ này đc ko, mk ko hiểu
⇔−4,5x2+72x+162=0⇔−4,5x2+72x+162=0
⇔(x−18)(−4,5x−9)=0
19 tháng 8 2021
1: Ta có: \(\dfrac{x+2}{x-2}+\dfrac{2}{x+2}=\dfrac{x^2}{x^2-4}\)
Suy ra: \(x^2+4x+4+2x-4=x^2\)
\(\Leftrightarrow6x=0\)
hay \(x=0\left(nhận\right)\)
2: Ta có: \(\dfrac{1}{x-6}-\dfrac{2}{x+6}=\dfrac{3x+6}{x^2-36}\)
Suy ra: \(x+6-2x+12=3x+6\)
\(\Leftrightarrow-x-3x=6-18=-12\)
hay \(x=3\left(nhận\right)\)
AH
Akai Haruma
Giáo viên
19 tháng 8 2021
Lời giải:
1. ĐKXĐ: $x\neq \pm 2$
PT \(\Leftrightarrow \frac{(x+2)^2+2(x-2)}{(x-2)(x+2)}=\frac{x^2}{x^2-4}\)
\(\Leftrightarrow \frac{x^2+6x}{x^2-4}=\frac{x^2}{x^2-4}\)
\(\Rightarrow x^2+6x=x^2\Leftrightarrow x=0\) (tm)
2. ĐKXĐ: $x\neq \pm 6$
PT \(\Leftrightarrow \frac{6+x-2(x-6)}{(x-6)(6+x)}=\frac{3x+6}{x^2-36}\)
\(\Leftrightarrow \frac{18-x}{x^2-36}=\frac{3x+6}{x^2-36}\)
\(\Rightarrow 18-x=3x+6\Leftrightarrow 12=4x\Leftrightarrow x=3\) (tm)
DC
3
27 tháng 2 2021
`(x/(x+1))^2+(x/(x-1))^2=90(x ne -1,1)`
`<=>x^2/(x+1)^2+x^2/(x-1)^2=90`
`<=>x^2(x-1)^2+x^2(x-1)^2=90(x^2-1)^2`
`<=>x^2(2x^2+2)=90(x^4-2x^2+1)`
`<=>2x^4+2x^2=90x^4-180x^2+90`
`<=>88x^4-182x^2+90=0`
`<=>88x^4-110x^2-72x^2+90=0`
`<=>22x^2(4x^2-5)-18(4x^2-5)=0`
`<=>(4x^2-5)(22x^2-18)=0`
`<=>(4x^2-5)(11x^2-9)=0`
`<=>` $\left[ \begin{array}{l}4x^2=5\\11x^2=9\end{array} \right.$
`<=>` $\left[ \begin{array}{l}x=\sqrt{\dfrac{5}{4}}\\x=-\sqrt{\dfrac{5}{4}}\\x=\sqrt{\dfrac{9}{11}}\\x=-\sqrt{\dfrac{9}{11}}\end{array} \right.$
Vậy `S={\sqrt{9/11},-\sqrt{9/11},\sqrt{5/4},-\sqrt{5/4}}`
A
27 tháng 2 2021
\(\left(\dfrac{x}{x+1}\right)^2+\left(\dfrac{x}{x-1}\right)^2=90\)
\(\Leftrightarrow\dfrac{x^2}{\left(x+1\right)^2}+\dfrac{x^2}{\left(x-1\right)^2}=90\)
\(\Leftrightarrow\dfrac{x^2\left(x-1\right)^2}{\left(x+1\right)^2\left(x-1\right)^2}+\dfrac{x^2\left(x+1\right)^2}{\left(x+1\right)^2\left(x-1\right)^2}=90\)
\(\Leftrightarrow\dfrac{x^2\left(x-1\right)^2+x^2\left(x+1\right)^2-90\left(x-1\right)^2\left(x+1\right)^2}{\left(x+1\right)^2\left(x-1\right)^2}=0\)
\(\Rightarrow x^2\left(x^2-2x+1\right)+x^2\left(x^2+2x+1\right)-90\left(x^2-1\right)^2=0\)
\(\Leftrightarrow x^4-2x^3+x^2+x^4+2x^3+x^2-90x^4+90x^2-90=0\)
\(\Leftrightarrow-88x^4+92x^2-90=0\)
27 tháng 1 2022
\(\Leftrightarrow\dfrac{4\cdot90\cdot\left(x+5\right)-4\cdot90\cdot x}{4x\left(x+5\right)}=\dfrac{x\left(x+5\right)}{4x\left(x+5\right)}\)
\(\Leftrightarrow x^2+5x-1800=0\)
\(\text{Δ}=5^2-4\cdot1\cdot\left(-1800\right)=7225>0\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{-5-85}{2}=\dfrac{-90}{2}=-45\left(nhận\right)\\x_2=\dfrac{-5+85}{2}=40\left(nhận\right)\end{matrix}\right.\)
9 tháng 3 2022
\(\Leftrightarrow\dfrac{x-5}{95}-1+\dfrac{x-132}{32}+1=\dfrac{x-131}{31}+1+\dfrac{x-10}{90}-1\)
=>x-100=0
hay x=100
TM
1
17 tháng 6 2023
=>x^2-4+x^2-3x=x^2+6
=>x^2-3x-4=6
=>x^2-3x-10=0
=>(x-5)(x+2)=0
=>x=5(nhận) hoặc x=-2(loại)
22 tháng 2 2019
\(\dfrac{x-90}{10}+\dfrac{x-76}{12}=\dfrac{x-58}{14}+\dfrac{x-36}{16}+\dfrac{x-15}{17}=15\)
\(\Leftrightarrow\left(\dfrac{x-90}{10}-1\right)+\left(\dfrac{x-76}{12}-2\right)=\left(\dfrac{x-58}{14}-3\right)+\left(\dfrac{x-36}{16}-4\right)+\left(\dfrac{x-15}{17}-5\right)\)\(\Leftrightarrow\dfrac{x-100}{10}+\dfrac{x-100}{12}=\dfrac{x-100}{14}+\dfrac{x-100}{16}+\dfrac{x-100}{17}\)
\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{10}+\dfrac{1}{12}-\dfrac{1}{14}-\dfrac{1}{16}-\dfrac{1}{17}\right)=0\)
\(\Leftrightarrow x-100=0\)
\(\Rightarrow x=100\)
Vậy \(S=\left\{100\right\}\)
\(\dfrac{90}{x}-\dfrac{36}{x-6}=2\) ( x # 0 ; x # 6)
⇔ \(\dfrac{90\left(x-6\right)-36x}{x\left(x-6\right)}=\dfrac{2x\left(x-6\right)}{x\left(x-6\right)}\)
⇔ 90x - 540 - 36x = 2x2 - 12x
⇔-2x2 + 66x - 540 = 0
⇔ -2( x2 - 33x +270 ) = 0
⇔ x2 - 18x - 15x + 270 = 0
⇔ x( x - 18) - 15( x - 18) = 0
⇔ ( x - 18)( x - 15) = 0
⇔ x = 18 ( TM) hoac x = 15 ( TM)
KL........