a] 2x+1/9 = 5/x+1
b] 2x-1/21 = 3/2x+1
c] 2x-1/2=5/x
d] x-3/3=2x+1/5
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25 tháng 8 2020
a) \(2x+\frac{3}{15}=\frac{7}{5}\)
=> \(2x=\frac{7}{5}-\frac{3}{15}=\frac{21}{15}-\frac{3}{15}=\frac{18}{15}\)
=> \(x=\frac{18}{15}:2=\frac{18}{15}\cdot\frac{1}{2}=\frac{9}{15}\cdot\frac{1}{1}=\frac{9}{15}\)
b) \(x-\frac{2}{9}=\frac{8}{3}\)
=> \(x=\frac{8}{3}+\frac{2}{9}\)
=> \(x=\frac{24}{9}+\frac{2}{9}=\frac{26}{9}\)
c) \(\frac{-8}{x}=\frac{-x}{18}\)
=> x(-x) = (-8).18
=> -x2 = -144
=> x2 = 144(bỏ dấu âm)
=> x = \(\pm\)12
d) \(\frac{2x+3}{6}=\frac{x-2}{5}\)
=> 5(2x + 3) = 6(x - 2)
=> 10x + 15 = 6x - 12
=> 10x + 15 - 6x + 12 = 0
=> 4x + 27 = 0
=> 4x = -27
=> x = -27/4
e) \(\frac{x+1}{22}=\frac{6}{x}\)
=> x(x + 1) = 132
=> x(x + 1) = 11.12
=> x = 11
f) \(\frac{2x-1}{2}=\frac{5}{x}\)
=> x(2x - 1) = 10
=> 2x2 - x = 10
=> 2x2 - x - 10 = 0
tới đây tự làm đi nhé
g) \(\frac{2x-1}{21}=\frac{3}{2x+1}\)
=> (2x - 1)(2x + 1) = 63
=> 4x2 - 1 = 63
=> 4x2 = 64
=> x2 = 16
=> x = \(\pm\)4
h) Tương tự
NC
25 tháng 8 2020
a) \(\frac{2x+3}{15}=\frac{7}{5}\Leftrightarrow10x+15=105\Leftrightarrow10x=90\Rightarrow x=9\)
b) \(\frac{x-2}{9}=\frac{8}{3}\Leftrightarrow3x-6=72\Leftrightarrow3x=78\Rightarrow x=26\)
c) \(\frac{-8}{x}=\frac{-x}{18}\Leftrightarrow x^2=144\Leftrightarrow\orbr{\begin{cases}x=12\\x=-12\end{cases}}\)
d) \(\frac{2x+3}{6}=\frac{x-2}{5}\Leftrightarrow10x+15=12x-12\Leftrightarrow2x=27\Rightarrow x=\frac{27}{2}\)
e) \(\frac{x+1}{22}=\frac{6}{x}\Leftrightarrow x^2+x-132=0\Leftrightarrow\left(x-11\right)\left(x+12\right)=0\Leftrightarrow\orbr{\begin{cases}x=11\\x=-12\end{cases}}\)
f) \(\frac{2x-1}{2}=\frac{5}{x}\Leftrightarrow2x^2-x-10=0\Leftrightarrow\left(x-2\right)\left(2x+5\right)=0\Leftrightarrow\orbr{\begin{cases}x=2\\x=-\frac{5}{2}\end{cases}}\)
g) \(\frac{2x-1}{21}=\frac{3}{2x+1}\Leftrightarrow4x^2=64\Leftrightarrow x^2=16\Rightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
h) \(\frac{10x+5}{6}=\frac{5}{x+1}\Leftrightarrow10x^2+15x-25=0\Leftrightarrow5\left(x-1\right)\left(2x+5\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{5}{2}\end{cases}}\)
24 tháng 2 2020
a) Ta có: \(\left(x+5\right)\left(2x-1\right)=\left(2x-3\right)\left(x+1\right)\)
\(\Leftrightarrow\left(x+5\right)\left(2x-1\right)-\left(2x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow2x^2-x+10x-5-\left(2x^2+2x-3x-3\right)=0\)
\(\Leftrightarrow2x^2+9x-5-2x^2+x+3=0\)
\(\Leftrightarrow10x-2=0\)
hay 10x=2
\(\Leftrightarrow x=\frac{1}{5}\)
Vậy: \(x=\frac{1}{5}\)
b) Ta có: \(\left(x+1\right)\left(x+9\right)=\left(x+3\right)\left(x+5\right)\)
\(\Leftrightarrow x^2+9x+x+9=x^2+5x+3x+15\)
\(\Leftrightarrow x^2+10x+9-x^2-8x-15=0\)
\(\Leftrightarrow2x-6=0\)
hay 2x=6
\(\Leftrightarrow x=3\)
Vậy: x=3
c) Ta có: \(\left(3x+5\right)\left(2x+1\right)=\left(6x-2\right)\left(x-3\right)\)
\(\Leftrightarrow6x^2+3x+10x+5=6x^2-18x-2x+6\)
\(\Leftrightarrow6x^2+13x+5=6x^2-20x+6\)
\(\Leftrightarrow6x^2+13x+5-6x^2+20x-6=0\)
\(\Leftrightarrow33x-1=0\)
\(\Leftrightarrow33x=1\)
hay \(x=\frac{1}{33}\)
Vậy: \(x=\frac{1}{33}\)
d) Ta có: \(\left(x-2\right)\left(3x+5\right)=\left(2x-4\right)\left(x+1\right)\)
\(\Leftrightarrow3x^2+5x-6x-10=2x^2+2x-4x-4\)
\(\Leftrightarrow3x^2-x-10=2x^2-2x-4\)
\(\Leftrightarrow3x^2-x-10-2x^2+2x+4=0\)
\(\Leftrightarrow x^2+x-6=0\)
\(\Leftrightarrow x^2+3x-2x-6=0\)
\(\Leftrightarrow x\left(x+3\right)-2\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{-3;2\right\}\)
đ) Ta có: \(9x^2-1=\left(3x+1\right)\left(2x-3\right)\)
\(\Leftrightarrow\left(3x-1\right)\left(3x+1\right)-\left(3x+1\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left[\left(3x-1\right)-\left(2x-3\right)\right]=0\)
\(\Leftrightarrow\left(3x+1\right)\left(3x-1-2x+3\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-1\\x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\frac{1}{3}\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{-\frac{1}{3};-2\right\}\)
e) Ta có: \(\left(2x+5\right)\left(x-4\right)=\left(x-5\right)\left(4-x\right)\)
\(\Leftrightarrow\left(2x+5\right)\left(x-4\right)+\left(x-5\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(2x+5+x-5\right)=0\)
\(\Leftrightarrow\left(x-4\right)\cdot3x=0\)
Vì \(3\ne0\)
nên \(\left[{}\begin{matrix}x-4=0\\x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=0\end{matrix}\right.\)
Vậy: \(x\in\left\{0;4\right\}\)
AH
Akai Haruma
Giáo viên
24 tháng 2 2020
a) $(x+5)(2x-1)=(2x-3)(x+1)$
$\Leftrightarrow 2x^2+9x-5=2x^2-x-3$
$\Leftrightarrow 10x=2\Rightarrow x=\frac{1}{5}$
b)
$(x+1)(x+9)=(x+3)(x+5)$
$\Leftrightarrow x^2+10x+9=x^2+8x+15$
$\Leftrightarrow 2x=6\Rightarrow x=3$
c)
$(3x+5)(2x+1)=(6x-2)(x-3)$
$\Leftrightarrow 6x^2+13x+5=6x^2-20x+6$
$\Leftrightarrow 33x=1\Rightarrow x=\frac{1}{33}$
10 tháng 9 2020
a) \(\left(2x-5\right)^2-\left(2x+3\right)\left(2x-3\right)=10\Leftrightarrow\left(4x^2-20x+25\right)-\left(4x^2-9\right)-10=0\)
\(\Leftrightarrow-20x+24=0\Leftrightarrow x=\frac{6}{5}\)
b) \(\left(4x-1\right)\left(x+2\right)-\left(2x+3\right)^2-5\left(x-1\right)=9\Leftrightarrow-10x-15=0\)
\(\Leftrightarrow x=\frac{-3}{2}\)
c) \(\left(x+1\right)^3-\left(x-1\right)^3-2=6\Leftrightarrow\left(x^3+3x^2+3x+1\right)-\left(x^3-3x^2+3x-1\right)-8=0\)
\(\Leftrightarrow6x^2-6=0\Leftrightarrow x=\pm1\)
d) \(\left(x+2\right)\left(x^2-2x+4\right)-\left(x+1\right)\left(x^2-x+1\right)-3\left(-x-2\right)=5\)
\(\Leftrightarrow\left(x^3+8\right)-\left(x^3+1\right)+3x+6=5\Leftrightarrow3x+8=0\Leftrightarrow x=\frac{-8}{3}\)
15 tháng 6 2023
`@` `\text {Ans}`
`\downarrow`
\(\dfrac{x-3}{3}=\dfrac{2x+1}{5}\)
`=> (x-3)5 = (2x+1)3`
`=> 5x-15 = 6x+3`
`=> 5x-6x = 15+3`
`=> -x=18`
`=> x=-18`
\(\dfrac{x+1}{22}=\dfrac{6}{x}\)
`=> (x+1)x = 22*6`
`=> (x+1)x = 132`
`=> x^2 + x = 132`
`=> x^2+x-132=0`
`=> (x-11)(x+12)=0`
`=>`\(\left[{}\begin{matrix}x-11=0\\x+12=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=11\\x=-12\end{matrix}\right.\)
\(\dfrac{2x-1}{2}=\dfrac{5}{x}\)
`=> (2x-1)x = 2*5`
`=> 2x^2 - x =10`
`=> 2x^2 - x - 10 =0`
`=> 2x^2 + 4x - 5x - 10 =0`
`=> (2x^2 + 4x) - (5x+10)=0`
`=> 2x(x+2) - 5(x+2)=0`
`=> (2x-5)(x+2)=0`
`=>`\(\left[{}\begin{matrix}2x-5=0\\x+2=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}2x=5\\x=-2\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-2\end{matrix}\right.\)
\(\dfrac{2x-1}{21}=\dfrac{3}{2x+1}\)
`=> (2x-1)(2x+1)=21*3`
`=> 4x^2 + 2x - 2x - 1 = 63`
`=> 4x^2 - 1=63`
`=> 4x^2 - 1 - 63=0`
`=> 4x^2 - 64 = 0`
`=> 4(x^2 - 16)=0`
`=> 4(x^2 + 4x - 4x - 16)=0`
`=> 4[(x^2+4x)-(4x+16)]=0`
`=> 4[x(x+4)-4(x+4)]=0`
`=> 4(x-4)(x+4)=0`
`=>`\(\left[{}\begin{matrix}x-4=0\\x+4=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
\(\dfrac{2x+1}{9}=\dfrac{5}{x+1}\)
`=> (2x+1)(x+1) = 9*5`
`=> (2x+1)(x+1)=45`
`=> 2x^2 + 2x + x + 1 = 45`
`=> 2x^2 + 3x + 1 =45`
`=> 2x^2 + 3x + 1 - 45 =0`
`=> 2x^2+3x-44=0`
`=> 2x^2 + 11x - 8x - 44=0`
`=> (2x^2 +11x) - (8x+44)=0`
`=> x(2x+11) - 4(2x+11)=0`
`=> (x-4)(2x+11)=0`
`=>`\(\left[{}\begin{matrix}x-4=0\\2x+11=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=4\\2x=-11\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=4\\x=-\dfrac{11}{2}\end{matrix}\right.\)
BD
15 tháng 6 2023
\(\dfrac{x-3}{3}=\dfrac{2x+1}{5}\\ \left(x-3\right)\cdot5=\left(2x+1\right)\cdot3\\ x5-15=6x+3\\ x5-6x=3+15\\ -x=18\\ \Rightarrow x=-18\)
\(\dfrac{x+1}{22}=\dfrac{6}{x}\\ \left(x+1\right)\cdot x=6\cdot22\\ \left(x+1\right)\cdot x=2\cdot3\cdot2\cdot11\\ \left(x+1\right)\cdot x=12\cdot11\\ \Rightarrow x=11\)
\(\dfrac{2x-1}{21}=\dfrac{3}{2x+1}\\ \left(2x-1\right)\cdot\left(2x+1\right)=21\cdot3\\ \left(2x-1\right)\cdot\left(2x+1\right)=7\cdot3\cdot3\\ \left(2x-1\right)\cdot\left(2x+1\right)=7\cdot9\\ \Rightarrow2x+1=9\\ 2x=8\\ x=4\)
30 tháng 6 2018
a) | 5/4x -7/2| - | 5/8x + 3/5| = 0
|5/4x - 7/2| = | 5/8x + 3/5|
TH1: 5/4x - 7/2 = 5/8x + 3/5
=> 5/4x - 5/8x = 3/5 +7/2
5/8x = 41/10
x = 41/10:5/8
x = 164/25
TH2: 5/4x - 7/2 = -5/8x - 3/5
=> 5/4x + 5/8x = -3/5 +7/2
15/8x = 29/10
x = 29/10 : 15/8
x = 116/75
KL: x = 164/25 hoặc x = 116/75
các bài cn lại b lm tương tự nha! h lm dài lắm!
Q
0
Q
1
18 tháng 8 2021
a: Ta có: \(\left|\dfrac{2}{5}-x\right|+\dfrac{1}{2}=3.5\)
\(\Leftrightarrow\left|x-\dfrac{2}{5}\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{2}{5}=3\\x-\dfrac{2}{5}=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{17}{5}\\x=-\dfrac{13}{5}\end{matrix}\right.\)
b: Ta có: \(\dfrac{21}{5}+3:\left|\dfrac{x}{4}-\dfrac{2}{3}\right|=6\)
\(\Leftrightarrow3:\left|\dfrac{1}{4}x-\dfrac{2}{3}\right|=6-\dfrac{21}{5}=\dfrac{9}{5}\)
\(\Leftrightarrow\left|\dfrac{1}{4}x-\dfrac{2}{3}\right|=\dfrac{5}{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{4}x-\dfrac{2}{3}=\dfrac{5}{3}\\\dfrac{1}{4}x-\dfrac{2}{3}=-\dfrac{5}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{4}x=\dfrac{7}{3}\\\dfrac{1}{4}x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{28}{3}\\x=-4\end{matrix}\right.\)
18 tháng 8 2021
em muốn hỏi là tại sao 3,5 bên trên xuống dưới lại là 3 và -x +2/5 của em xuống dưới lại chuyển thành x-2/5 ạ mong anh giải đáp
2 tháng 7 2018
(x+2)(x+3)-(x-2)(x+5)=0
=> x2+5x+6-x2-3x+10=0
=>2x+16=0
=>2x=-16
=>x=-8
a) \(\dfrac{2x+1}{9}=\dfrac{5}{x+1}\left(x\ne-1\right)\)
\(\Rightarrow\left(2x+1\right)\left(x+1\right)=9\cdot5=45\)
\(\Rightarrow2x^2+2x+x+1=45\)
\(\Rightarrow2x^2+3x-44=0\)
\(\Rightarrow2x^2+11x-8x-44=0\)
\(\Rightarrow x\left(2x+11\right)-4\left(2x+11\right)=0\)
\(\Rightarrow\left(x-4\right)\left(2x+11\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\2x=-11\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{11}{2}\end{matrix}\right.\)
b) \(\dfrac{2x-1}{21}=\dfrac{3}{2x+1}\left(x\ne-\dfrac{1}{2}\right)\)
\(\Rightarrow\left(2x-1\right)\left(2x+1\right)=21\cdot3=63\)
\(\Rightarrow4x^2-1=63\)
\(\Rightarrow4x^2=64\)
\(\Rightarrow\left(2x\right)^2=8^2\)
\(\Rightarrow\left[{}\begin{matrix}2x=8\\2x=-8\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
c) \(\dfrac{2x-1}{2}=\dfrac{5}{x}\left(x\ne0\right)\)
\(\Rightarrow x\left(2x-1\right)=5\cdot2=10\)
\(\Rightarrow2x^2-x=10\)
\(\Rightarrow2x^2-x-10=0\)
\(\Rightarrow2x^2+4x-5x-10=0\)
\(\Rightarrow2x\left(x+2\right)-5\left(x+2\right)=0\)
\(\Rightarrow\left(2x-5\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x=5\\x=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-2\end{matrix}\right.\)
d) \(\dfrac{x-3}{3}=\dfrac{2x+1}{5}\)
\(\Rightarrow15\cdot\dfrac{x-3}{3}=15\cdot\dfrac{2x+1}{5}\)
\(\Rightarrow5\left(x-3\right)=3\left(2x+1\right)\)
\(\Rightarrow5x-15=6x+3\)
\(\Rightarrow6x-5x=-18\)
\(\Rightarrow x=-18\)