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a) (2x - 3)(6 - 2x) = 0
=> \(\left[{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.=>\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=3\end{matrix}\right.\)
b) \(5\dfrac{4}{7}:x=13=>\dfrac{39}{7}:x=13=>x=\dfrac{39}{7}:13=>x=\dfrac{3}{7}\)
c) \(2x-\dfrac{3}{7}=6\dfrac{2}{7}=>2x-\dfrac{3}{7}=\dfrac{44}{7}=>2x=\dfrac{47}{7}=>x=\dfrac{47}{14}\)
d) \(\dfrac{x}{5}+\dfrac{1}{2}=\dfrac{6}{10}=>\dfrac{x}{5}=\dfrac{6}{10}-\dfrac{1}{2}=>\dfrac{x}{5}=\dfrac{1}{10}=>x.10=5=>x=\dfrac{1}{2}\)
e) \(\dfrac{x+3}{15}=\dfrac{1}{3}=>\left(x+3\right).3=15=>x+3=5=>x=2\)
4 câu đầu hìn như sai đề :v
`m)(3/2-2/(-5)):x-1/2=3/2`
`<=>(3/2+2/5):x=3/2+1/2=2`
`<=>19/10:x=2`
`<=>x=19/10:2=19/20`
`n)(3/2-5/11-3/13)(2x-2)=(-3/4+5/22+3/26)`
`<=>(3/2-5/11-3/13)(2x-2)+3/4-5/22-3/26=0`
`<=>(3/2-5/11-3/13)(2x-2)+1/2(3/2-5/11-3/13)=0`
`<=>(3/2-5/11-3/13)(2x-2+1/2)=0`
Mà `3/2-5/11-3/13>0`
`<=>2x-2+1/2=0`
`<=>2x-3/2=0`
`<=>2x=3/2<=>x=3/4`
1) PT \(\Leftrightarrow\left(\dfrac{x+1}{35}+1\right)+\left(\dfrac{x+3}{33}+1\right)=\left(\dfrac{x+5}{31}+1\right)+\left(\dfrac{x+7}{29}+1\right)\)
\(\Leftrightarrow\dfrac{x+36}{35}+\dfrac{x+36}{33}=\dfrac{x+36}{31}+\dfrac{x+36}{29}\)
\(\Leftrightarrow\left(x+36\right)\left(\dfrac{1}{29}+\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{35}\right)=0\)
\(\Leftrightarrow x+36=0\) (Do \(\dfrac{1}{29}+\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{35}>0\))
\(\Leftrightarrow x=-36\).
Vậy nghiệm của pt là x = -36.
a: \(A=\dfrac{-7}{28}\cdot\dfrac{15}{25}=\dfrac{-1}{4}\cdot\dfrac{3}{5}=\dfrac{-3}{20}\)
b: \(B=\dfrac{-5\cdot7}{14\cdot\left(-3\right)}=\dfrac{35}{42}=\dfrac{5}{6}\)
c: \(C=\dfrac{-1}{5}-\dfrac{1}{5}\cdot\dfrac{3}{5}=\dfrac{-1}{5}-\dfrac{3}{25}=\dfrac{-8}{25}\)
d: \(D=\dfrac{-3}{4}-\dfrac{1}{4}=-1\)
e: \(E=\dfrac{-4}{5}\left(1-\dfrac{15}{16}\right)=\dfrac{-4}{5}\cdot\dfrac{1}{16}=\dfrac{-1}{20}\)
f: \(F=\dfrac{6-7}{4}\cdot\dfrac{4+12}{22}=\dfrac{-1}{4}\cdot\dfrac{8}{11}=\dfrac{-2}{11}\)
`h)x/2-1/x=1/12(x ne 0)`
`<=>6x^2-12=x`
`<=>6x^2-x-12=0`
`<=>6x^2-9x+8x-12=0`
`<=>3x(2x-3)+4(2x-3)=0`
`<=>(2x-3)(3x+4)=0`
`<=>` \(\left[ \begin{array}{l}x=\dfrac32\\x=-\dfrac43\end{array} \right.\)
`i)x^2-7/6x+1/3=0`
`<=>6x^2-7x+2=0`
`<=>6x^2-3x-4x+2=0`
`<=>3x(2x-1)-2(2x-1)=0`
`<=>(2x-1)(3x-2)=0`
`<=>` \(\left[ \begin{array}{l}x=\dfrac12\\x=\dfrac23\end{array} \right.\)
Câu cuối không có dấu "=" nên không tìm được x :v
- Hai câu h, i bấm nốt đáp án để đẹp nha ;-; câu k thiếu đề :v
a: =>1/2x=7/2-2/3=21/6-4/6=17/6
=>x=17/3
b: =>2/3:x=-7-1/3=-22/3
=>x=2/3:(-22/3)=-1/11
c: =>1/3x+2/5x-2/5=0
=>11/15x=2/5
hay x=6/11
d: =>2x-3=0 hoặc 6-2x=0
=>x=3/2 hoặc x=3
\(a,\dfrac{5}{13}\times\dfrac{4}{15}\times13=\dfrac{5\times4\times13}{13\times5\times3}=\dfrac{4}{3}\\ b,\left(\dfrac{3}{7}+\dfrac{5}{2}\right)\times\dfrac{7}{5}=\dfrac{3}{7}\times\dfrac{7}{5}+\dfrac{5}{2}\times\dfrac{7}{5}=\dfrac{3}{5}+\dfrac{7}{2}=\dfrac{6}{10}+\dfrac{35}{10}=\dfrac{41}{10}\\ c,\dfrac{1}{5}\times\dfrac{11}{18}+\dfrac{11}{18}\times\dfrac{3}{5}=\dfrac{11}{18}\times\left(\dfrac{1}{5}+\dfrac{3}{5}\right)=\dfrac{11}{18}\times\dfrac{4}{5}=\dfrac{22}{45}\)
a: \(\Leftrightarrow\dfrac{7x+10}{x+1}\left(x^2-x-2-2x^2+3x+5\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(-x^2+2x+3\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\cdot\left(x^2-2x-3\right)=0\)
=>(7x+10)(x-3)=0
=>x=3 hoặc x=-10/7
b: \(\Leftrightarrow\dfrac{13}{\left(2x+7\right)\left(x-3\right)}+\dfrac{1}{2x+7}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow13\left(x+3\right)+x^2-9-12x-42=0\)
\(\Leftrightarrow x^2-12x-51+13x+39=0\)
\(\Leftrightarrow x^2+x-12=0\)
=>(x+4)(x-3)=0
=>x=-4
a) \(\dfrac{x+3}{x-3}-\dfrac{x-3}{x+3}=\dfrac{36}{x^2-9}\)
\(\Rightarrow\dfrac{\left(x+3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{\left(x-3\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{36}{\left(x-3\right)\left(x+3\right)}\)
\(\Rightarrow\dfrac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}-\dfrac{\left(x-3\right)^2}{\left(x-3\right)\left(x+3\right)}=\dfrac{36}{\left(x-3\right)\left(x+3\right)}\)
\(\Rightarrow\dfrac{\left(x+3\right)^2-\left(x-3\right)^2}{\left(x-3\right)\left(x+3\right)}=\dfrac{36}{\left(x-3\right)\left(x+3\right)}\)
\(\Rightarrow\left(x+3\right)^2-\left(x-3\right)^2=36\)
\(\Rightarrow\left(x^2+6x+9\right)-\left(x^2-6x+9\right)=36\)
\(\Rightarrow x^2+6x+9-x^2+6x-9=36\)
\(\Rightarrow12x=36\)
\(\Rightarrow x=\dfrac{36}{12}\)
Vậy x = 3
b) \(x^2-x-6=0\)
\(\Rightarrow x^2-3x+2x-6=0\)
\(\Rightarrow x\left(x-3\right)+2\left(x-3\right)=0\)
\(\Rightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
c) \(\dfrac{2x-1}{5}-\dfrac{x-2}{3}=\dfrac{x+17}{15}\)
\(\Rightarrow\dfrac{3\left(2x-1\right)}{15}-\dfrac{5\left(x-2\right)}{15}=\dfrac{x+17}{15}\)
\(\Rightarrow\dfrac{3\left(2x-1\right)-5\left(x-2\right)}{15}=\dfrac{x+17}{15}\)
\(\Rightarrow\dfrac{6x-3-5x+10}{15}=\dfrac{x+17}{15}\)
... Phần còn lại cũng tương tự như vậy thôi
e: =>2/7-x=2/5
=>7-x=5
=>x=2
f: =>2x+3/3=10/3
=>2x+3=10
=>2x=7
=>x=7/2
g: =>(14+x)/7=15/7
=>x+14=15
=>x=1
h: =>(2x+3)/x=13/x
=>2x+3=13
=>2x=10
=>x=5