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a: \(\Leftrightarrow15\left(x-1\right)-2\left(7x+3\right)\le10\left(2x+1\right)+6\left(3-2x\right)\)
\(\Leftrightarrow15x-15-14x-6\le20x+10+18-12x\)
=>x-21<=8x+28
=>-7x<=49
hay x>=-7
b: \(\Leftrightarrow20\left(2x+1\right)-15\left(2x^2+3\right)< 10x\left(5-3x\right)-12\left(4x+1\right)\)
\(\Leftrightarrow40x+20-30x^2-45< 50x-30x^2-48x-12\)
=>40x-25<2x-12
=>38x<13
hay x<13/38
\(a,\dfrac{x-1}{2}-\dfrac{7x+3}{15}\le\dfrac{2x+1}{3}+\dfrac{3-2x}{5}\\ \Leftrightarrow\dfrac{15\left(x-1\right)}{30}-\dfrac{2\left(7x+3\right)}{30}\le\dfrac{10\left(2x+1\right)}{30}+\dfrac{6\left(3-2x\right)}{30}\\ \Leftrightarrow15x-15-14x-6\le20x+10+18-12x\\ \Leftrightarrow x-21\le8x+28\\ \Leftrightarrow7x+49\ge0\\ \Leftrightarrow x\ge-7\)
\(b,\dfrac{2x+1}{-3}-\dfrac{2x^2+3}{-4}>\dfrac{x\left(5-3x\right)}{-6}-\dfrac{4x+1}{-5}\\ \Leftrightarrow\dfrac{20\left(2x+1\right)}{-60}-\dfrac{15\left(2x^2+3\right)}{-60}>\dfrac{10x\left(5-3x\right)}{-60}-\dfrac{12\left(4x+1\right)}{-60}\\ \Leftrightarrow40x+20-30x^2-45>50x-30x^2-48x-12\\ \Leftrightarrow38x-13>0\\ \Leftrightarrow x>\dfrac{13}{38}\)
\(1,\dfrac{4x-4}{3}=\dfrac{7-x}{5}\\ \Leftrightarrow5\left(4x-4\right)=3\left(7-x\right)\\ \Leftrightarrow20x-20=21-3x\\ \Leftrightarrow17x=41\Leftrightarrow x=\dfrac{41}{17}\)
\(2,\dfrac{3x-9}{5}=\dfrac{3-x}{2}\\ \Leftrightarrow6x-18=15-5x\\ \Leftrightarrow11x=33\\ \Leftrightarrow x=3\)
\(3,\dfrac{2x-1}{5}-\dfrac{3-x}{3}=1\\ \Leftrightarrow\dfrac{6x-3-15+5x}{15}=1\\ \Leftrightarrow11x-18=1\\ \Leftrightarrow x=\dfrac{19}{11}\)
\(4,\dfrac{x-5}{3}+\dfrac{3x+4}{2}=\dfrac{5x+2}{6}\\ \Leftrightarrow2x-10+9x+12=5x+2\\ \Leftrightarrow6x=0\Leftrightarrow x=0\)
\(5,\dfrac{x-3}{2}+\dfrac{2x+3}{5}=\dfrac{2x+5}{10}\\ \Leftrightarrow5x-15+4x+6=2x+5\\ \Leftrightarrow7x=14\\ \Leftrightarrow x=2\)
Tick nha
2: Ta có: \(\dfrac{3x-9}{5}=\dfrac{3-x}{2}\)
\(\Leftrightarrow6x-18=15-5x\)
\(\Leftrightarrow11x=33\)
hay x=3
\(\dfrac{2x-1}{3}+\dfrac{3x-2}{4}=\dfrac{4x-3}{5}\)
\(\Leftrightarrow\dfrac{20\left(2x-1\right)}{60}+\dfrac{15\left(3x-2\right)}{60}=\dfrac{12\left(4x-3\right)}{60}\)
`<=> 20(2x-1) +15(3x-2) =12(4x-3)`
`<=> 40x - 20 + 45x - 30 = 48x - 36`
`<=> 85x -50 = 48x - 36`
`<=> 85x-48x = -36+50`
`<=> 37x =14`
`<=> x= 14/37`
Vậy phương trình có nghiệm `x=14/37`
__
\(\dfrac{5}{x-3}+\dfrac{4}{x+3}=\dfrac{x-6}{x^2-9}\)
\(\Leftrightarrow\dfrac{5}{x-3}+\dfrac{4}{x+3}=\dfrac{x-6}{\left(x-3\right)\left(x+3\right)}\)
ĐKXĐ : \(\left\{{}\begin{matrix}x-3\ne0\\x+3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x\ne-3\end{matrix}\right.\)
Ta có : \(\dfrac{5}{x-3}+\dfrac{4}{x+3}=\dfrac{x-6}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow\dfrac{5\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{4\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{x-6}{\left(x-3\right)\left(x+3\right)}\)
`=> 5x + 15 + 4x -12=x-6`
`<=> 9x + 3=x-6`
`<=> 9x-x=-6-3`
`<=> 8x = -9`
`<=>x=-9/8(tm)`
Vậy phương trình có nghiệm `x=-9/8`
` @ yngoc`
\(1,\dfrac{4x-3}{x-5}=\dfrac{29}{3}\left(ĐKXĐ:x\ne5\right)\)
\(\Rightarrow3\left(4x-3\right)=29\left(x-5\right)\)
\(\Leftrightarrow12x-9=29x-145\)
\(\Leftrightarrow12x-9-29x+145=0\)
\(\Leftrightarrow-17x+136=0\)
\(\Leftrightarrow-17x=-136\)
\(\Leftrightarrow x=8\left(tm\right)\)
Vậy \(S=\left\{8\right\}\)
\(2,\dfrac{2x-1}{5-3x}=2\left(ĐKXĐ:x\ne\dfrac{5}{3}\right)\)
\(\Rightarrow2x-1=2\left(5-3x\right)\)
\(\Leftrightarrow2x-1=10-6x\)
\(\Leftrightarrow2x-1-10+6x=0\)
\(\Leftrightarrow8x-11=0\)
\(\Leftrightarrow8x=11\)
\(\Leftrightarrow x=\dfrac{11}{8}\left(tm\right)\)
Vậy \(S=\left\{\dfrac{11}{8}\right\}\)
\(3,\dfrac{4x-5}{x-1}=2+\dfrac{x}{x-1}\left(ĐKXĐ:x\ne1\right)\)
\(\Leftrightarrow\dfrac{4x-5}{x-1}=\dfrac{2\left(x-1\right)}{x-1}+\dfrac{x}{x-1}\)
\(\Leftrightarrow\dfrac{4x-5}{x-1}=\dfrac{2x-2}{x-1}+\dfrac{x}{x-1}\)
\(\Leftrightarrow\dfrac{4x-5}{x-1}=\dfrac{3x-2}{x-1}\)
\(\Rightarrow4x-5=3x-2\)
\(\Leftrightarrow4x-5-3x+2=0\)
\(\Leftrightarrow x-3=0\)
\(\Leftrightarrow x=3\left(tm\right)\)
Vậy \(S=\left\{3\right\}\)
\(4,\dfrac{2x+5}{2x}-\dfrac{x}{x+5}=0\left(ĐKXĐ:x\ne\dfrac{1}{2};x\ne-5\right)\)
\(\Leftrightarrow\dfrac{\left(2x+5\right)\left(x+5\right)}{2x\left(x+5\right)}-\dfrac{2x^2}{2x\left(x+5\right)}=0\)
\(\Leftrightarrow\dfrac{2x^2+15x+25}{2x\left(x+5\right)}-\dfrac{2x^2}{2x\left(x+5\right)}=0\)
\(\Leftrightarrow\dfrac{15x+25}{2x\left(x+5\right)}=0\)
\(\Rightarrow15x+25=0\)
\(\Leftrightarrow15x=-25\)
\(\Leftrightarrow x=\dfrac{-5}{3}\left(tm\right)\)
Vậy \(S=\left\{\dfrac{-5}{3}\right\}\)
\(1,\dfrac{4x-3}{x-5}=\dfrac{29}{3}\)
\(\Leftrightarrow\dfrac{3\left(4x-3\right)-29\left(x-5\right)}{3\left(x-5\right)}=0\)
\(\Leftrightarrow12x-9-29x+145=0\)
\(\Leftrightarrow-17x=-136\)
\(\Leftrightarrow x=8\)
\(2,\dfrac{2x-1}{5-3x}=2\)
\(\Leftrightarrow\dfrac{2x-1-2\left(5-3x\right)}{5-3x}=0\)
\(\Leftrightarrow2x-1-10+6x=0\)
\(\Leftrightarrow8x=11\)
\(\Leftrightarrow x=\dfrac{11}{8}\)
\(3,\dfrac{4x-5}{x-1}=2+\dfrac{x}{x-1}\)
\(\Leftrightarrow\dfrac{4x-5-2\left(x-1-x\right)}{x-1}=0\)
\(\Leftrightarrow4x-5-2x+2+2x=0\)
\(\Leftrightarrow4x=3\)
\(\Leftrightarrow x=\dfrac{3}{4}\)
\(4,\dfrac{2x+5}{2x}-\dfrac{x}{x+5}=0\)
\(\Leftrightarrow\dfrac{\left(2x+5\right)\left(x+5\right)-2x^2}{2x\left(x+5\right)}=0\)
\(\Leftrightarrow2x^2+10x+5x+25-2x^2=0\)
\(\Leftrightarrow15x=-25\)
\(\Leftrightarrow x=-\dfrac{5}{3}\)
TK
https://lazi.vn/edu/exercise/giai-phuong-trinh-4x-5-x-1-2-x-x-1-7-x-2-3-x-5
a: \(\Leftrightarrow4x-5=2x-2+x\)
=>4x-5=3x-2
=>x=3(nhận)
b: =>7x-35=3x+6
=>4x=41
hay x=41/4(nhận)
c: \(\Leftrightarrow\dfrac{14}{3\left(x-4\right)}-\dfrac{x+2}{x-4}=\dfrac{-3}{2\left(x-4\right)}-\dfrac{5}{6}\)
\(\Leftrightarrow\dfrac{28}{6\left(x-4\right)}-\dfrac{6\left(x+2\right)}{6\left(x-4\right)}=\dfrac{-9}{6\left(x-4\right)}-\dfrac{5\left(x-4\right)}{6\left(x-4\right)}\)
\(\Leftrightarrow28-6x-12=-9-5x+20\)
=>-6x+16=-5x+11
=>-x=-5
hay x=5(nhận)
d: \(\Leftrightarrow x^2+2x+1-\left(x^2-2x+1\right)=16\)
\(\Leftrightarrow4x=16\)
hay x=4(nhận)
1a.
ĐKXĐ: \(x\ne\left\{1;3\right\}\)
\(\Leftrightarrow\dfrac{6}{x-1}=\dfrac{4}{x-3}+\dfrac{4}{x-3}\)
\(\Leftrightarrow\dfrac{3}{x-1}=\dfrac{4}{x-3}\Leftrightarrow3\left(x-3\right)=4\left(x-1\right)\)
\(\Leftrightarrow3x-9=4x-4\Rightarrow x=-5\)
b.
ĐKXĐ: \(x\ne\left\{-1;2\right\}\)
\(\Leftrightarrow\dfrac{5}{x+1}=\dfrac{3}{2-x}+\dfrac{1}{2-x}\)
\(\Leftrightarrow\dfrac{5}{x+1}=\dfrac{4}{2-x}\Leftrightarrow5\left(2-x\right)=4\left(x+1\right)\)
\(\Leftrightarrow10-2x=4x+4\Leftrightarrow6x=6\Rightarrow x=1\)
1c.
ĐKXĐ: \(x\ne\left\{2;5\right\}\)
\(\Leftrightarrow\dfrac{3x\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}-\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x-5\right)}=\dfrac{-3x}{\left(x-2\right)\left(x-5\right)}\)
\(\Leftrightarrow3x\left(x-5\right)-x\left(x-2\right)=-3x\)
\(\Leftrightarrow2x^2-10x=0\Leftrightarrow2x\left(x-5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=5\left(loại\right)\end{matrix}\right.\)
2a.
\(\Leftrightarrow-4x^2-5x+6=x^2+4x+4\)
\(\Leftrightarrow5x^2+9x-2=0\Rightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{5}\end{matrix}\right.\)
2b.
\(2x^2-6x+1=0\Rightarrow x=\dfrac{3\pm\sqrt{7}}{2}\)
`a,` \(\dfrac{5x+2}{6}-\dfrac{8x-1}{3}=\dfrac{4x+2}{5}-5\)
`<=> (5(5x+2))/30 - (10(8x-1))/30 = (6(4x+2))/30 - (5.30)/30`
`<=> 5(5x+2) - 10(8x-1) =6(4x+2) - 5.30`
`<=> 25x + 10 - 80x + 10 = 24x+12 - 150`
`<=> -55x +20 = 24x-138`
`<=> -55x -24x=-138-20`
`<=>-79x=-158`
`<=> x=2`
Vậy pt có nghiệm `x=2`
`b,` \(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x\left(x-2\right)}\)
ĐKXĐ : \(\left\{{}\begin{matrix}x-2\ne0\\x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne2\\x\ne0\end{matrix}\right.\)
Ta có : `(x+2)/(x-2) -1/x = 2/(x(x-2))`
`<=> (x(x+2))/(x(x-2)) - (x-2)/(x(x-2)) = 2/(x(x-2))`
`=> x^2 +2x - x +2 = 2`
`<=> x^2 + x =0`
`<=>x(x+1)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(l\right)\\x=-1\end{matrix}\right.\)
Vậy pt có nghiệm `x=-1`
`c,2x^3 + 6x^2 =x^2 +3x`
`<=> 2x^3 + 6x^2 -x^2 -3x=0`
`<=> 2x^3 + 5x^2 -3x=0`
`->` Đề có sai ko ạ ?
`d,` \(\left|x-4\right|+3x=5\) `(1)`
Thường hợp `1` : `x-4 >= 0<=> x >=0` thì phương trình `(1)` thở thành :
`x-4 = 5-3x`
`<=> x+3x=5+4`
`<=> 4x=9`
`<=> x= 9/4 (t//m)`
Trường hợp `2` : `x-4< 0<=> x<0` thì phương trình `(1)` trở thành :
`-(x-4) =5-3x`
`<=> -x +4=5-3x`
`<=> -x+3x=5-4`
`<=> 2x =1`
`<=>x=1/2 ( kt//m)`
Vậy phương trình có nghiệm `x=9/4`
