giúp mik với, 12 + 2 x 2 = .... + 25
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(- 47 ) - 53 = (- 47 ) + ( - 53 ) = -100
(- 43 ) - ( -43 ) =0
Ta có : -(4+7) =-11
( -4 ) - 7 = -11
Vì - ( 4+7 ) = -11 , ( -4 ) - 7 =-11 nên -(4 +7) = ( - 4) -7
Ta có : -( 12 -25 ) = 13
( -12 + 25 )=13
Vì -(12 -25 ) = 13; (-12 +25) = 13 nên - (12 - 25 ) = ( -12 +25 )
\(25\left(x+y\right)^2-16\left(x-y\right)^2\)
\(=\left(5x+5y\right)^2-\left(4x-4y\right)^2\)
\(=\left(5x+5y+4x-4y\right)\left(5x+5y-4x+4y\right)\)
\(=\left(9x+y\right)\left(x+9y\right)\)
\(\left(x-1\right)^2=5^2\\\Rightarrow x-1=5\\ \Rightarrow x=5+1=6\)
=>\(5\cdot\dfrac{3\sqrt{x-3}}{5}-7\cdot\dfrac{2\sqrt{x-3}}{3}-7\cdot\sqrt{x^2-9}+18\cdot\sqrt{\dfrac{9}{81}\left(x^2-9\right)}=0\)
=>\(3\cdot\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}=7\cdot\sqrt{x^2-9}-18\cdot\dfrac{3}{9}\cdot\sqrt{x^2-9}\)
=>\(-\dfrac{5}{3}\sqrt{x-3}=\sqrt{x^2-9}\)
=>\(\sqrt{x-3}\left(\sqrt{x+3}+\dfrac{5}{3}\right)=0\)
=>x-3=0
=>x=3
\(\dfrac{-2}{9}\) . \(\dfrac{5}{2}\) + \(\dfrac{25}{12}\)
= \(\dfrac{-10}{18}\) + \(\dfrac{25}{12}\)
= \(\dfrac{165}{108}\)
\(\left(x+\frac{1}{5}\right)^2=\frac{26}{25}-\frac{17}{25}\)
\(\left(x+\frac{1}{5}\right)^2=\frac{9}{25}\)
\(\left(x+\frac{1}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(x+\frac{1}{5}=\frac{3}{5}\)
\(x=\frac{3}{5}-\frac{1}{5}\)
\(x=\frac{2}{5}\)
vậy \(x=\frac{2}{5}\)
\(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)
\(\left(x+\frac{1}{5}\right)^2=\frac{26}{25}-\frac{17}{25}\)
\(\left(x+\frac{1}{5}\right)^2=\frac{9}{25}\)
\(\left(x+\frac{1}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(x+\frac{1}{5}=\frac{3}{5}\)
\(x=\frac{3}{5}-\frac{1}{5}\)
\(x=\frac{2}{5}\)
\(x^2-10x=-25\)
\(x^2-10x+25=0\)
\(\left(x-5\right)^2=0\)
\(x-5=0\)
\(x=5\)
\(x^2-10x=-25\)
\(\Leftrightarrow x^2-10x+25=0\)
\(\Leftrightarrow\left(x-5\right)^2=0\)
\(\Leftrightarrow x=5\)
12+2 nhân 2= 3+25 nhé
chỗ ... là 3