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\(\dfrac{3x+2}{5x-7}=\dfrac{3x-1}{5x+1}\)
\(\Leftrightarrow\left(3x+2\right)\left(5x+1\right)=\left(3x-1\right)\left(5x-7\right)\)
\(\Leftrightarrow15x^2+3x+10x+2=15x^2-21x-5x-7\)
\(\Leftrightarrow15x^2+13x+2=15x^2-26x-7\)
\(\Leftrightarrow15x^2-15x^2-13x-2=26x-7\)
\(\Leftrightarrow-13x-2=26x-7\)
\(\Leftrightarrow26x+13x=7+2\)
\(\Leftrightarrow39x=9\Leftrightarrow x=\dfrac{3}{13}\)
b tương tự
1: \(\Leftrightarrow\left(x+1\right)^2=4\)
=>x+1=2 hoặc x+1=-2
=>x=1 hoặc x=-3
2: \(\Leftrightarrow7x-21=5x+25\)
=>2x=46
=>x=23
3: \(\Leftrightarrow x^2+4x+3=x^2+0.5x+4x+2\)
=>4,5x+2=4x+3
=>x=1
a: \(\dfrac{3x+2}{5x+7}=\dfrac{3x-1}{5x+1}\)
\(\Leftrightarrow\left(3x+2\right)\left(5x+1\right)=\left(3x-1\right)\left(5x+7\right)\)
\(\Leftrightarrow15x^2+3x+10x+2=15x^2+21x-5x-7\)
=>16x-7=13x+2
=>3x=9
hay x=3
b: \(\dfrac{x+1}{2016}+\dfrac{x}{2017}=\dfrac{x+2}{2015}+\dfrac{x+3}{2014}\)
\(\Leftrightarrow\left(\dfrac{x+1}{2016}+1\right)+\left(\dfrac{x}{2017}+1\right)=\left(\dfrac{x+2}{2015}+1\right)+\left(\dfrac{x+3}{2014}+1\right)\)
=>x+2017=0
hay x=-2017
e: \(\left(2x-3\right)^2=144\)
=>2x-3=12 hoặc 2x-3=-12
=>2x=15 hoặc 2x=-9
=>x=15/2 hoặc x=-9/2
a: =1/2x^3*x^2-1/2x^3*6x-1/2x^3*10
=1/2x^5-3x^4-5x^3
b: =-3x^2*5x^3+3x^2*4x^2-3x^2*3x+3x^2*3x
=-15x^5+12x^4-9x^3+9x^2
c: \(=3x\cdot5x^2-3x\cdot2x-3x=15x^3-6x^2-3x\)
d: \(=\dfrac{1}{2}x^2y\cdot2x^3-\dfrac{1}{2}x^2y\cdot\dfrac{2}{5}xy^2-\dfrac{1}{2}x^2y=x^5y-\dfrac{1}{5}x^3y^3-\dfrac{1}{2}x^2y\)
a: \(\dfrac{31-2x}{x+23}=\dfrac{9}{4}\)
=>121-8x=9x+207
=>-17x=86
hay x=-86/17
b: \(\dfrac{\left|2x-1\right|}{\dfrac{1}{2}}=\dfrac{18}{5}\)
=>|2x-1|=9/5
=>2x-1=9/5 hoặc 2x-1=-9/5
=>2x=14/5 hoặc 2x=-4/5
=>x=7/5 hoặc x=-2/5
a) Áp dụng tính chất dãy tỉ số bằng nhau , ta có :
\(\dfrac{3x+2}{5x+7}=\dfrac{3x-1}{5x+1}=\dfrac{3x+2-3x+1}{5x+7-5x-1}=\dfrac{3}{6}=\dfrac{1}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}3x+2=\dfrac{5x+7}{2}\\3x-1=\dfrac{5x+1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x+4=5x+7\\6x-2=5x+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=3\end{matrix}\right.\)(Nhận)
Vậy x = 3
b) Giống vậy :)
a, \(\dfrac{3x+2}{5x+7}=\dfrac{3x-1}{5x+1}\)
\(\Leftrightarrow\left(3x+2\right)\left(5x+1\right)=\left(5x+7\right)\left(3x-1\right)\)
\(\Leftrightarrow15x^2+3x+10x+2=15x^2-5x+21x-7\)
\(\Leftrightarrow15x^2+13x=15x^2+16x-9\)
\(\Leftrightarrow3x=9\)
\(\Leftrightarrow x=3\)
Vậy x = 3
b, Tương tự
a/ đk: x khác -7/5 ; x khác -1/5
pt <=> \(\dfrac{\left(3x+2\right)\left(5x+1\right)}{\left(5x+7\right)\left(5x+1\right)}=\dfrac{\left(3x-1\right)\left(5x+7\right)}{\left(5x+7\right)\left(5x+1\right)}\)
\(\Rightarrow15x^2+13x+2=15x^2+16x-7\)
\(\Leftrightarrow15x^2+13x-15x^2-16x^2=-7-2\)
\(\Leftrightarrow-3x=-9\Leftrightarrow x=3\left(tm\right)\)
vậy x = 3
b/ đk: x khác -1/2; x khác -3
pt <=> \(\dfrac{\left(x+1\right)\left(x+3\right)}{\left(2x+1\right)\left(x+3\right)}=\dfrac{\left(0,5x+2\right)\left(2x+1\right)}{\left(2x+1\right)\left(x+3\right)}\)
\(\Rightarrow x^2+4x+3=x^2+4,5x+2\)
\(\Leftrightarrow x^2+4x-x^2-4,5x=2-3\)
\(\Leftrightarrow-0,5x=-1\Leftrightarrow x=2\left(tm\right)\)
vậy x = 2
a) Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{3x+2}{5x+7}=\dfrac{3x-1}{5x+1}=\dfrac{\left(3x+2\right)-\left(3x-1\right)}{\left(5x+7\right)-\left(5x+1\right)}=\dfrac{3}{6}=\dfrac{1}{2}\)
\(\Rightarrow2\left(3x+2\right)=5x+7\)
\(\Rightarrow6x+4=5x+7\)
\(\Leftrightarrow x=3\)
Vậy x = 3
b) Ta có: \(\dfrac{0,5x+2}{x+3}=\dfrac{2\left(0,5x+2\right)}{2\left(x+3\right)}=\dfrac{x+4}{2x+6}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x+1}{2x+1}=\dfrac{0,5x+2}{x+3}=\dfrac{x+4}{2x+6}=\dfrac{\left(x+4\right)-\left(x+1\right)}{\left(2x+6\right)-\left(2x+1\right)}=\dfrac{3}{5}\)
\(\Rightarrow5\left(x+1\right)=3\left(2x+1\right)\)
\(\Rightarrow5x+5=6x+3\)
\(\Leftrightarrow x=2\)