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Lời giải:
PT $\Leftrightarrow \frac{11}{17}x+\frac{3}{17}=\frac{78}{17}x-\frac{22}{17}$
$\Leftrightarrow \frac{25}{17}=\frac{67}{17}x$
$x=\frac{25}{67}$
\(x-\dfrac{3}{17}\left(2x-1\right)=\dfrac{7}{34}\left(1-2x\right)+\dfrac{10x-3}{2}\)
<=>\(x-\dfrac{3}{17}\left(2x-1\right)-\dfrac{7}{34}\left(1-2x\right)=\dfrac{10x-3}{2}\)
<=>\(x-\dfrac{3}{17}\left(2x-1\right)+\dfrac{7}{34}\left(2x-1\right)=\dfrac{10x-3}{2}\)
<=>\(x+\left(-\dfrac{3}{17}+\dfrac{7}{34}\right)\left(2x-1\right)-\dfrac{10x-3}{2}=0\)
<=>\(x+\dfrac{1}{34}\left(2x-1\right)-\dfrac{10x-3}{2}=0\)
<=>\(x+\dfrac{1}{17}x-\dfrac{1}{34}-\dfrac{10x-3}{2}=0\)
<=>\(\dfrac{18}{17}x-\dfrac{85x-25.5}{17}=\dfrac{0.5}{17}\)
=> \(18x-85x+25.5=0.5\)
<=>\(-67x=0.5-25.5\)
<=>\(-67x=-25\)
<=> \(x=\dfrac{25}{67}\)
đây là cách làm khác (dù dài hơn) nếu bạn không hiểu cách bên dưới
chúc bạn học tốt!
a: \(\Leftrightarrow7\left(7-3x\right)+12\left(5x+2\right)=84\left(x+13\right)\)
\(\Leftrightarrow49-21x+60x+24=84x+1092\)
\(\Leftrightarrow39x-84x=1092-73\)
=>-45x=1019
hay x=-1019/45
b: \(\Leftrightarrow21\left(x+3\right)-14=4\left(5x+9\right)-7\left(7x-9\right)\)
=>21x+63-14=20x+36-49x+63
=>21x+49=-29x+99
=>50x=50
hay x=1
c: \(\Leftrightarrow7\left(2x+1\right)-3\left(5x+2\right)=21x+63\)
=>14x+7-15x-6-21x-63=0
=>-22x-64=0
hay x=-32/11
d: \(\Leftrightarrow35\left(2x-3\right)-15\left(2x+3\right)=21\left(4x+3\right)-17\cdot105\)
=>70x-105-30x-45=84x+63-1785
=>40x-150-84x+1722=0
=>-44x+1572=0
hay x=393/11
a: =>\(\dfrac{5x-15+4x-8}{\left(x-2\right)\left(x-3\right)}=\dfrac{1}{x}\)
=>\(\dfrac{9x-23}{\left(x-2\right)\left(x-3\right)}=\dfrac{1}{x}\)
=>9x^2-23x=x^2-5x+6
=>8x^2-18x-6=0
=>\(x=\dfrac{9\pm\sqrt{129}}{8}\)
b: =>\(\dfrac{12x+1}{11x-4}=\dfrac{20x+17-20x+8}{18}=\dfrac{25}{18}\)
=>216x+18=275x-100
=>-59x=-118
=>x=2
a) \(\dfrac{2-x}{3}-x-2\le\dfrac{x-17}{2}\) \(\Leftrightarrow\) \(6\left(\dfrac{2-x}{3}-x-2\right)\le6\left(\dfrac{x-17}{2}\right)\) \(\Leftrightarrow\) 4-2x-6x-12\(\le\)3x-51 \(\Leftrightarrow\) -2x-6x-3x\(\le\)-51-4+12 \(\Leftrightarrow\) -11x\(\le\)-43 \(\Rightarrow\) x\(\ge\)43/11.
b) \(\dfrac{2x+1}{3}-\dfrac{x-4}{4}\le\dfrac{3x+1}{6}-\dfrac{x-4}{12}\) \(\Leftrightarrow\) \(12\left(\dfrac{2x+1}{3}+\dfrac{4-x}{4}\right)\le12\left(\dfrac{3x+1}{6}+\dfrac{4-x}{12}\right)\) \(\Leftrightarrow\) 8x+4+12-3x\(\le\)6x+2+4-x \(\Leftrightarrow\) 8x-3x-6x+x\(\le\)2+4-4-12 \(\Leftrightarrow\) 0x\(\le\)-10 (vô lí).
a) \(\dfrac{2-x}{3}-x-2\le\dfrac{x-17}{2}\)
\(\Leftrightarrow2\left(2-x\right)-6\left(x+2\right)\le3\left(x-17\right)\)
\(\Leftrightarrow4-2x-6x-12\le3x-51\)
\(\Leftrightarrow-11x\le-43\)
\(\Leftrightarrow x\ge\dfrac{43}{11}\)
Vậy S = {\(x\) | \(x\ge\dfrac{43}{11}\) }
b) \(\dfrac{2x+1}{3}-\dfrac{x-4}{4}\le\dfrac{3x+1}{6}-\dfrac{x-4}{12}\)
\(\Leftrightarrow4\left(2x+1\right)-3\left(x-4\right)\le2\left(3x+1\right)-\left(x-4\right)\)
\(\Leftrightarrow8x+4-3x+12\le6x+2-x+4\)
\(\Leftrightarrow0x\le-10\) (vô lý)
Vậy \(S=\varnothing\)
f: =>\(\dfrac{14}{3\left(x-4\right)}-\dfrac{x+2}{x-4}=\dfrac{-3}{2\left(x-4\right)}-\dfrac{5}{6}\)
=>28-6(x+2)=-9-5(x-4)
=>28-6x-12=-9-5x+20
=>-6x+16=-5x+11
=>-x=-5
=>x=5
d: =>\(\dfrac{12x+1}{11x-4}=\dfrac{20x+17-20x+8}{18}=\dfrac{25}{18}\)
=>25(11x-4)=18(12x+1)
=>275x-100=216x+18
=>59x=118
=>x=2
f: =>\(\dfrac{14}{3\left(x-4\right)}-\dfrac{x+2}{x-4}=\dfrac{-3}{2\left(x-4\right)}-\dfrac{5}{6}\)
=>28-6(x+2)=-9-5(x-4)
=>28-6x-12=-9-5x+20
=>-6x+16=-5x+11
=>-x=-5
=>x=5
\(B=\dfrac{11}{17}+\left(-\dfrac{8}{19}\right)+\left(-\dfrac{3}{4}\right)+\dfrac{6}{17}-\dfrac{30}{19}\\ \Rightarrow B=\left(\dfrac{11}{17}+\dfrac{6}{17}\right)+\left(-\dfrac{8}{19}-\dfrac{30}{19}\right)-\dfrac{3}{4}\\ \Rightarrow B=1-2-\dfrac{3}{4}\\ \Rightarrow B=-\dfrac{7}{4}\)
bạn tách một câu vài câu hỏi chứ đừng gộp như thế này ko ai trả lời đâu
a: =>\(4x-5=2x-2+x=3x-2\)
=>x=3
b: \(\Leftrightarrow7x-35=3x+6\)
=>4x=41
=>x=41/4
c: =>(2x+5)(x+5)-2x^2=0
=>2x^2+10x+5x+25-2x^2=0
=>15x=-25
=>x=-5/3
e: \(\Leftrightarrow\dfrac{11}{x}=\dfrac{9x-36+2x+2}{\left(x+1\right)\left(x-4\right)}\)
=>11(x^2-3x-4)=x(11x-34)
=>11x^2-33x-44=11x^2-34x
=>x=44
Ta có: \(\dfrac{-1}{3}\cdot\dfrac{141}{17}-\dfrac{39}{3}\cdot\dfrac{-1}{17}\)
\(=\dfrac{-1}{3}\cdot\dfrac{141}{17}-\dfrac{-1}{3}\cdot\dfrac{39}{17}\)
\(=\dfrac{-1}{3}\cdot\left(\dfrac{141}{17}-\dfrac{39}{17}\right)\)
\(=\dfrac{-1}{3}\cdot6=-2\)