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Đề thế này hả e
\(\dfrac{2x-y}{x+y}=\dfrac{2}{3}\)
\(\Leftrightarrow3\left(2x-y\right)=2\left(x+y\right)\)
\(\Leftrightarrow6x-3y=2x+y\)
\(\Leftrightarrow4x=4y\)
\(\Leftrightarrow x=y\)
Vậy.....
\(\dfrac{2x-y}{x+y}=\dfrac{2}{3}\Leftrightarrow3\left(2x-y\right)=2\left(x+y\right)\)
\(\Leftrightarrow6x-3y=2x+2y\)
\(\Leftrightarrow4x=5y\)
\(\Leftrightarrow\dfrac{x}{y}=\dfrac{5}{4}\)
Vậy....
a làm lại nhé, nãy sai
5x(x-3)2 - 5(x-1)3 + 15(x+4) (x-4) = 5
<=> 5x( x2 - 2. x .3 + 32 ) -5 ( x3 -3.x2.1 + 3.x.12 - 13 ) + 15. ( x2 - 42 ) = 5
<=>5x( x2 - 6x + 9 ) - 5 ( x3 - 3x2 + 3x - 1 ) + 15. ( x2 - 16 ) =5
<=> 5x.x2 + 5x.( - 6x ) + 5x.9 - 5.x3 -5. ( -3x2 ) - 5.3x - 5.(-1) +15.x2 + 15. ( -16 ) = 5
<=> 5x3 - 30x2 + 45x - 5x3 + 15x2 - 15x + 5 + 15x2 - 240 = 5
<=> 5x3 - 5x3 - 30x2 + 15x2 + 15x2 + 45x - 15x + 5 - 240 = 5
<=> 30x - 235 = 5
<=> 30x = 5 + 235
<=> 30x = 240
<=> x = 240 : 30
<=> x = 8
a, \(\left(x^2-9\right)^2-\left(x-3\right)\left(x+3\right)\left(x^2+9\right)=\left(x^2-9\right)^2-\left(x^2-9\right)\left(x^2+9\right)\)
\(=x^4-18x^2+81-x^4+81=-18x^2+162\)
b, \(\left(x^2+x-3\right)\left(x^2-x+3\right)=\left[x^4-\left(x-3\right)^2\right]\)
\(=x^4-x^2+6x-9\)
a) \(\left(2x-3\right)^2-\left(2x+5\right)^2=10\)
\(\Leftrightarrow4x^2-12x+9-4x^2-20x-25-10=0\)
\(\Leftrightarrow-32x-26=0\)
\(\Leftrightarrow-32x=26\)
\(\Rightarrow x=-\frac{13}{16}\)
b) \(4\left(x+1\right)^2+\left(2x-1\right)^2+8\left(x-1\right)\left(x+1\right)=11\)
\(\Leftrightarrow4x^2+8x+4+4x^2-4x+1+8x^2-8=0\)
\(\Leftrightarrow16x^2+4x-3=0\)
\(\Leftrightarrow4\left(4x^2+x+\frac{1}{16}\right)-\frac{13}{4}=0\)
\(\Leftrightarrow\left[2\left(2x+\frac{1}{4}\right)\right]^2-\left(\frac{\sqrt{13}}{2}\right)^2=0\)
\(\Leftrightarrow\left(4x+\frac{1}{2}-\frac{\sqrt{13}}{2}\right)\left(4x+\frac{1}{2}+\frac{\sqrt{13}}{2}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}4x+\frac{1-\sqrt{13}}{2}=0\\4x+\frac{1+\sqrt{13}}{2}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{\sqrt{13}-1}{8}\\x=\frac{-1-\sqrt{13}}{8}\end{cases}}\)
c) \(\left(x+5\right)^2=45+x^2\)
\(\Leftrightarrow x^2+10x+25-x^2-45=0\)
\(\Leftrightarrow10x-20=0\)
\(\Leftrightarrow10x=20\)
\(\Rightarrow x=2\)
d) \(\left(2x-3\right)^2-\left(2x-1\right)^2=-3\)
\(\Leftrightarrow4x^2-12x+9-4x^2+4x-1+3=0\)
\(\Leftrightarrow-8x+11=0\)
\(\Leftrightarrow-8x=-11\)
\(\Rightarrow x=\frac{11}{8}\)
e) \(\left(x-1\right)^2-\left(5x-3\right)^2=0\)
\(\Leftrightarrow\left(x-1-5x+3\right)\left(x-1+5x-3\right)=0\)
\(\Leftrightarrow\left(-4x+2\right)\left(6x-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}-4x+2=0\\6x-4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{2}{3}\end{cases}}\)
a ) 2x ( x - 5 ) - x ( 3 + 2x ) = 26
2x2 - 10x - 3x - 2x2 = 26
- 13x = 26
x = 26 : ( -13 )
x = -2
b) 49x2 - 81 = 0
( 7x - 9 )( 7x + 9 ) = 0
Th1 :
7x - 9 = 0
7x = 9
x = \(\frac{9}{7}\)
Th2
7x + 9 = 0
7x = -9
x = \(-\frac{9}{7}\)
Vay x = \(\frac{9}{7}\) hoac x = \(-\frac{9}{7}\)
Ta có : (2x + 3).(x + 4) + (x - 5).(x - 2) = (3x - 5)(x - 4)
<=> 2x2 + 11x + 12 + x2 - 7x - 10 = 3x2 - 17x - 20
<=> 2x2 + 11x + 12 + x2 - 7x - 10 - 3x2 + 17x - 20 = 0
<=> 2x2 + x2 - 3x2 + 11x - 7x + 17x + 12 - 10 - 20 = 0
<=> 21x - 18 = 0
<=> 21x = 18
<=> x = \(\frac{18}{21}=\frac{6}{7}\)