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+)TH1:a=\(\frac{1}{3}\) và b=\(\frac{1}{4}\)
\(y=\frac{5}{3}.\frac{1}{3}-\frac{3}{\frac{1}{4}}=-\frac{67}{6}\)
+)TH2 a=\(-\frac{1}{3}\)và b=\(-\frac{1}{4}\)
\(y=\frac{5}{3}.-\frac{1}{3}-\frac{3}{-\frac{1}{4}}=\frac{103}{9}\)
\(A=\left|x-\dfrac{1}{2}\right|+\dfrac{3}{4}\\ \text{Do }\left|x-\dfrac{1}{2}\right|\ge0\forall x\\ A=\left|x-\dfrac{1}{2}\right|+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)
Dấu \("="\) xảy ra khi :
\(\left|x-\dfrac{1}{2}\right|=0\\ \Leftrightarrow x-\dfrac{1}{2}=0\\ \Leftrightarrow x=\dfrac{1}{2}\)
Vậy \(A_{\left(Min\right)}=\dfrac{3}{4}\) khi \(x=\dfrac{1}{2}\)
\(B=2-\left|x+\dfrac{5}{6}\right|\\ \text{Do }\left|x+\dfrac{5}{6}\right|\ge0\forall x\\ \Rightarrow B=2-\left|x+\dfrac{5}{6}\right|\le2\forall x\)
Dấu \("="\) xảy ra khi :
\(\left|x+\dfrac{5}{6}\right|=0\\ \Leftrightarrow x+\dfrac{5}{6}=0\\ \Leftrightarrow x=-\dfrac{5}{6}\)
Vậy \(B_{\left(Max\right)}=2\) khi \(x=-\dfrac{5}{6}\)
a, \(x^2+y^2=8\Rightarrow\left(x+y\right)^2-2xy=8\Rightarrow xy=\frac{8-\left(x+y\right)^2}{-2}=\frac{8-4}{-2}=-2\)
=>\(M=x^3+x^4+y^3+y^4=\left(x+y\right)^3-3xy\left(x+y\right)+\left(x^2+y^2\right)^2-2x^2y^2\)
\(=2^3-3.\left(-2\right).2+8^2-2.\left(-2\right)^2=76\)
b, \(M=x^2+y^2+2xy-4x-4y+3=\left(x+y\right)^2-4\left(x+y\right)+4-1=\left(x+y-2\right)^2-1=\left(5-2\right)^2-1=8\)
a)
Ta có:
\(2xy=(x+y)^2-(x^2+y^2)=2^2-8=-4\Rightarrow xy=-2\)
Vậy:
\(M=x^3+x^4+y^3+y^4=(x^3+y^3)+(x^4+y^4)\)
\(=(x+y)(x^2+y^2)-xy(x+y)+(x^2+y^2)^2-2x^2y^2\)
\(=2.8-(-2).2+8^2-2(-2)^2\)
\(=76\)
b)
\(M=x^2+y^2+2xy-4x-4y+3\)
\(=(x^2+xy)+(y^2+xy)-4(x+y)+3\)
\(=x(x+y)+y(x+y)-4(x+y)+3\)
\(=(x+y)(x+y)-4(x+y)+3\)
\(=5.5-4.5+3=8\)
a) \(\left|x\right|=2,1\)
x= +- 2,1
b) \(\left|x\right|=\frac{3}{4}\left(x< 0\right)\)
x= -3/4
c) \(\left|x\right|=-1\frac{2}{5}\)
\(x\in\varphi\)
d) \(\left|x\right|=0,35\left(x>0\right)\)
\(x=0,35\)
a) |x| = 2,1 <=> \(\orbr{\begin{cases}x=2,1\\x=-2,1\end{cases}}\)
b) |x| = 3/4 <=> x = - 3/4 ( do x < 0 )
c) ko tim dc x vi |x| >= 0 voi moi x
d) |x| = 0,35 <=> x = 0,35 ( do x>0 )