\(\dfrac{2x-1}{4}-\dfrac{x-5}{3}\le\dfrac{4x-1}{12}+2\)
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TT
21 tháng 1 2022
a) \(\dfrac{x+1}{4}-\dfrac{5+2x}{8}=\dfrac{3-4x}{2}\)
⇔\(\dfrac{2\left(x+1\right)}{8}-\dfrac{5+2x}{8}=\dfrac{4\left(3-4x\right)}{8}\)
⇔ 2x + 2 - 5 - 2x = 12 -16x
⇔ 16x = 15
⇔ x = 15/16
b) \(\dfrac{4-3x}{5}-\dfrac{4-x}{10}=\dfrac{x+2}{2}\)
⇔\(\dfrac{2\left(4-3x\right)}{10}-\dfrac{4-x}{10}=\dfrac{5\left(x+2\right)}{10}\)
⇔ 8 - 6x - 4 + x = 5x + 10
⇔ 10x = -6
⇔ x = -6/10
21 tháng 1 2022
Câu 1:
x + 1/4 - 5 + 2x/8 = 3 - 4x/2
<=> 2x + 2/8 - 5 + 2x/8 = 12 - 16x/8
<=> 2x + 2 - 5 - 2x = 12 - 16x
<=> -3 = 12 - 16x <=> 15 = 16x <=> x = 15/16
Câu 2:
4 - 3x/5 - 4 - x/10 = x + 2/2
<=> 8 - 6x/10 - 4 - x/10 = 5x + 10/10
<=> 8 - 6x - 4 + x = 5x + 10
<=> 4 - 5x = 5x + 10
<=> 4 = 10x + 10 <=> 10x = -6 <=> x = -3/5
30 tháng 3 2022
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
19 tháng 1 2022
Câu 1:
\(\left(x-2\right)\left(x^2+2x+4\right)+25x=x\left(x+5\right)\left(x-5\right)+8\)
\(\Leftrightarrow x^3-8+25x=x\left(x^2-25\right)+8\)
\(\Leftrightarrow x^3-8+25x=x^3-25x+8\)
\(\Leftrightarrow x^3-8+25x-x^3+25x-8=0\)
\(\Leftrightarrow50x-16=0\)
\(\Leftrightarrow50x=16\)
\(\Leftrightarrow x=\dfrac{8}{25}\)
B
19 tháng 1 2022
Câu 2 :
\(\dfrac{x+5}{4}+\dfrac{3+2x}{3}=\dfrac{6x-1}{3}-\dfrac{1-2x}{12}\)
<=> \(\dfrac{3\left(x+5\right)}{12}+\dfrac{4\left(3+2x\right)}{12}=\dfrac{4\left(6x-1\right)}{12}-\dfrac{1-2x}{12}\)
<=>\(\dfrac{3x+15+12+8x}{12}=\dfrac{24x-4-1+2x}{12}\)
<=> 3x + 15 + 12 + 8x = 24x - 4 - 1 +2x
<=> 11x+27 = 26x -5
<=> ( 26x - 5 ) - ( 11x + 27 ) = 0
<=> 15x - 32 = 0
<=> 15x = 32
<=> x = \(\dfrac{32}{15}\)
4 tháng 3 2022
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
4 tháng 3 2022
\(a,\dfrac{x-1}{2}-\dfrac{7x+3}{15}\le\dfrac{2x+1}{3}+\dfrac{3-2x}{5}\\ \Leftrightarrow\dfrac{15\left(x-1\right)}{30}-\dfrac{2\left(7x+3\right)}{30}\le\dfrac{10\left(2x+1\right)}{30}+\dfrac{6\left(3-2x\right)}{30}\\ \Leftrightarrow15x-15-14x-6\le20x+10+18-12x\\ \Leftrightarrow x-21\le8x+28\\ \Leftrightarrow7x+49\ge0\\ \Leftrightarrow x\ge-7\)
\(b,\dfrac{2x+1}{-3}-\dfrac{2x^2+3}{-4}>\dfrac{x\left(5-3x\right)}{-6}-\dfrac{4x+1}{-5}\\ \Leftrightarrow\dfrac{20\left(2x+1\right)}{-60}-\dfrac{15\left(2x^2+3\right)}{-60}>\dfrac{10x\left(5-3x\right)}{-60}-\dfrac{12\left(4x+1\right)}{-60}\\ \Leftrightarrow40x+20-30x^2-45>50x-30x^2-48x-12\\ \Leftrightarrow38x-13>0\\ \Leftrightarrow x>\dfrac{13}{38}\)
10 tháng 11 2021
\(a,ĐK:...\\ PT\Leftrightarrow x^2-6x=x^2-7x+10\\ \Leftrightarrow x=10\left(tm\right)\\ b,ĐK:...\\ PT\Leftrightarrow2x\left(4-x\right)-\left(2-2x\right)\left(8-x\right)=\left(8-x\right)\left(4-x\right)\\ \Leftrightarrow8x-2x^2+16+18x-2x^2=32-12x+x^2\\ \Leftrightarrow3x^2-38x+16=0\left(casio\right)\\ c,ĐK:...\\ PT\Leftrightarrow2x\left(x-4\right)-4x=0\\ \Leftrightarrow2x^2-12x=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=6\left(tm\right)\end{matrix}\right.\)
10 tháng 5 2023
g: =>12x+1>=36x+12-24x-3
=>12x+1>=12x+9(loại)
h: =>6(x-1)+4(2-x)<=3(3x-3)
=>6x-6+8-4x<=9x-9
=>2x+2<=9x-9
=>-7x<=-11
=>x>=11/7
i: =>4x^2-12x+9>4x^2-3x
=>-12x+9>-3x
=>-9x>-9
=>x<1
AH
Akai Haruma
Giáo viên
13 tháng 12 2021
Lời giải:
a.
\(\frac{10}{x+2}=\frac{60}{6(x+2)}=\frac{60(x-2)}{6(x+2)(x-2)}=\frac{60(x-2)}{6(x^2-4)}\)
\(\frac{5}{2x-4}=\frac{15(x+2)}{6(x-2)(x+2)}=\frac{15(x+2)}{6(x^2-4)}\)
\(\frac{1}{6-3x}=\frac{x+2}{3(2-x)}=\frac{2(x+2)^2}{6(2-x)(2+x)}=\frac{-2(x+2)^2}{6(x^2-4)}\)
b.
\(\frac{1}{x+2}=\frac{x(2-x)}{x(x+2)(2-x)}=\frac{x(2-x)}{x(4-x^2)}\)
\(\frac{8}{2x-x^2}=\frac{8(x+2)}{(x+2)x(2-x)}=\frac{8(x+2)}{x(4-x^2)}\)
c.
\(\frac{4x^2-3x+5}{x^3-1}\)
\(\frac{1-2x}{x^2+x+1}=\frac{(1-2x)(x-1)}{(x-1)(x^2+x+1)}=\frac{-2x^2+3x-1}{x^3-1}\)
\(-2=\frac{-2(x^3-1)}{x^3-1}\)
\(\Leftrightarrow6x-3-4x+20< =4x-1+24\)
=>2x+17-4x-23<=0
=>-2x-6<=0
=>-2x<=6
hay x>=-3
\(\Leftrightarrow\dfrac{6x-3-4x+20-4x+1-24}{12}\le0\)
\(\Rightarrow\left\{{}\begin{matrix}-2x-6< 0\\-2x-6\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>-3\\x\ne-3\end{matrix}\right.\)