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`(3x+2)/3 <= (x-4)/7`
`<=>7(3x+2) <= 3(x-4)`
`<=> 21x+14<=3x-12`
`<=>18x <= -26`
`<=> x <=-13/9`
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐK: ` x\ne \pm 3`
`(x+1)/(x-3)+(x-1)/(x+3)=(x+6)/(x^2-9)`
`<=>(x+1)(x+3)+(x-1)(x-3)=x+6`
`<=>x^2+4x+3+x^2-4x+3=x+6`
`<=>2x^2+6=x+6`
`<=>2x^2-x=0`
`<=>x(2x-1)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy `S={0; 1/2}`.
ĐKXĐ: x ≠ -3, x ≠ 3
\(\dfrac{x+1}{x-3}+\dfrac{x-1}{x+3}=\dfrac{x+6}{x^2-9}\)
\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+3\right)+\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{x+6}{\left(x-3\right)\left(x+3\right)}\)
\(\Rightarrow x^2+4x+3+x^2-4x+3=x+6\)
\(\Leftrightarrow2x^2-x=0\)
\(\Leftrightarrow x\left(2x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=\dfrac{1}{2}\left(tm\right)\end{matrix}\right.\)
Vậy...
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\(\left(2x-5\right).2=\left(x+2\right).3\)
\(\Rightarrow4x-10=3x+6\)
\(\Rightarrow x=16\)
\(\dfrac{2x+5}{3}=\dfrac{x+2}{2}\)
MTC : 6
Quy đồng mẫu thức :
\(\Rightarrow\) \(\dfrac{2\left(2x+5\right)}{6}=\)\(\dfrac{3\left(x+2\right)}{6}\)
Suy ra : 2(2x + 5) = 3(x + 2)
\(\Leftrightarrow\) 4x + 10 = 3x + 6
\(\Leftrightarrow\) 4x + 10 - 3x - 6 = 0
\(\Leftrightarrow\) x + 4 = 0
\(\Leftrightarrow\) x = - 4
Vậy S = \(\left\{-4\right\}\)
Chúc bạn học tốt
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a: \(\Leftrightarrow4\left(2x+1\right)-3\left(6x-1\right)=2x+1\)
=>8x+4-18x+3=2x+1
=>-10x+7=2x+1
=>-12x=-6
hay x=1/2
b: \(\Leftrightarrow4x^2-12x+7x-21-x^2=3x^2+6x\)
=>5x-21=6x
=>-x=21
hay x=-21
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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 \(S=\varnothing\)
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\(\text{a) }\dfrac{5x^2-3x}{5}+\dfrac{3x+1}{4}< \dfrac{x\left(2x+1\right)}{2}-\dfrac{3}{2}\\ \Leftrightarrow4\left(5x^2-3x\right)+5\left(3x+1\right)< 10x\left(2x+1\right)-15\\ \Leftrightarrow20x^2-12x+15x+5< 20x^2+10x-15\\ \Leftrightarrow20x^2+3x-20x^2-10x< -15-5\\ \Leftrightarrow-7x< -20\\ \Leftrightarrow x>\dfrac{20}{7}\)
Vậy bất phương trình có nghiệm \(x>\dfrac{20}{7}\)
\(\text{b) }\dfrac{5x-20}{3}-\dfrac{2x^2+x}{2}\ge\dfrac{x\left(1-3x\right)}{3}-\dfrac{5x}{4}\\ \Leftrightarrow4\left(5x-20\right)-6\left(2x^2+x\right)\ge4x\left(1-3x\right)-15x\\ \Leftrightarrow20x-80-12x^2-6x\ge4x-12x^2-15x\\ \Leftrightarrow-12x^2+14x+12x^2+11x\ge80\\ \Leftrightarrow25x\ge80\\ \Leftrightarrow x\ge\dfrac{16}{5}\)
Vậy bất phương trình có nghiệm \(x\ge\dfrac{16}{5}\)
\(\text{c) }\left(x+3\right)^2\le x^2-7\\ \Leftrightarrow x^2+6x+9\le x^2-7\\ \Leftrightarrow x^2+6x-x^2\le-7-9\\ \Leftrightarrow6x\le-16\\ \Leftrightarrow x\le-\dfrac{8}{3}\)
Vậy bất phương trình có nghiệm \(x\le-\dfrac{8}{3}\)
`(3x+2)/3 <= (x-4)/7`
`<=>7(3x+2) <= 3(x-4)`
`<=>21x+14 <= 3x-12`
`<=> 18x <=-26`
`<=>x <= -13/9`
Vậy `x<=-13/9`.