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*\(M+\left(5x^2-2xy\right)=6x^2+9xy-y^2\)
\(M=6x^2+9xy-y^2-\left(5x^2-2xy\right)\)
\(M=6x^2+9xy-y^2-5x^2+2xy\)
\(M=\left(6-5\right)x^2+\left(9+2\right)xy-y^2\)
\(M=x^2+11xy-y^2\)
* \(\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}\le0\)
Ta có : \(\hept{\begin{cases}\left(2x-5\right)^{2018}\ge0\forall x\\\left(3y+4\right)^{2020}\ge0\forall y\end{cases}\Rightarrow}\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}\ge0\forall x,y\)
Mà đề cho \(\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}\le0\)
=> \(\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}=0\)
=> \(\hept{\begin{cases}2x-5=0\\3y+4=0\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{5}{2}\\y=-\frac{4}{3}\end{cases}}\)
Thay x = 5/2 ; y = -4/3 vào M ta được :
\(M=\left(\frac{5}{2}\right)^2+11\cdot\frac{5}{2}\cdot\left(-\frac{4}{3}\right)-\left(-\frac{4}{3}\right)^2\)
\(M=\frac{25}{4}+\frac{-110}{3}-\frac{16}{9}\)
\(M=\frac{-1159}{36}\)
Vậy giá trị của M = -1159/36 khi x = 5/2 ; y = -4/3
Không chắc nha
Lời giải:
\(|2x-5|+x=21\)
\(\Leftrightarrow |2x-5|=21-x\)
\(\Rightarrow \left[\begin{matrix} 2x-5=21-x\\ 2x-5=x-21\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{26}{3}\\ x=-16\end{matrix}\right.\) (đều thỏa mãn)
Vậy \(x=\frac{26}{3}; x=-16\)
\(x^2-x+1=0\)
\(\Leftrightarrow\left(x-1\right)^2=0\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\)
Vậy/....
\(\frac{11x-2}{7x-5}=\frac{11}{8}\)
\(\Leftrightarrow8\left(11x-2\right)=11\left(7x-5\right)\)
\(\Leftrightarrow88x-16=77x-55\)
\(\Leftrightarrow88x-77x=-55+16\)
\(\Leftrightarrow11x=-39\)
\(\Leftrightarrow x=\frac{-39}{11}\)
\(\frac{12x-2}{7x-8}=\frac{11}{8}\)
\(\Rightarrow8\left(12x-2\right)=11\left(7x-8\right)\)
\(\Rightarrow96x-16=77x-88\)
\(\Rightarrow96x-77x=-88+16\)
\(\Rightarrow19x=-72\)
\(\Rightarrow x=-\frac{72}{19}\)
\(\left(\dfrac{1}{2}\right)^x=\left(\dfrac{1}{8}\right)^{x-2}\)
\(\Leftrightarrow\left(\dfrac{1}{2}\right)^x=\left(\dfrac{1}{2}\right)^{3x-6}\)
\(\Leftrightarrow x=3x-6\)
\(\Leftrightarrow3x-x=6\)
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=3\left(tm\right)\)
Vậy ........
\(\left(\dfrac{1}{2}\right)^x=\left(\dfrac{1}{8}\right)^{x-2}\\ \Rightarrow\left(\dfrac{1}{2}\right)^x=\left(\dfrac{1^3}{2^3}\right)^{x-2}\\ \Rightarrow\left(\dfrac{1}{2}\right)^x=\left(\dfrac{1}{2}\right)^{3\left(x-2\right)}\\ \Leftrightarrow3\left(x-2\right)=x\\ \Rightarrow3x-6=x\\ \Rightarrow3x-x=6\\ \Rightarrow x\left(3-1\right)=6\\ \Rightarrow2x=6\\ \Rightarrow x=6:2=3\)
\(\left|-\frac{2}{5}\right|\sqrt{16}-2\sqrt{\frac{25}{9}}+2017^0\)
= \(\frac{2}{5}.4-2.\frac{5}{3}+1\)
= \(\frac{8}{5}-\frac{10}{3}+1\)
= \(-\frac{11}{15}\)
a) f(-2) = 4.(-2) - 12 = (-8) - 12 = (-20)
f(3) = 4 . 3 - 12 = 13 - 12 = 1
b) y = 5 => 4x = 5 + 12 = 17
=> x = 17 : 4 = 17/4 = 4,25