Tim x biet
a) x-2 trên 5=3phan8
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a: \(\left(x+5\right)^2>=0\forall x\)
\(\left(2y-8\right)^2>=0\forall y\)
Do đó: \(\left(x+5\right)^2+\left(2y-8\right)^2>=0\forall x,y\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x+5=0\\2y-8=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=-5\\y=4\end{matrix}\right.\)
b: \(\left(x+3\right)\left(2y-1\right)=5\)
=>\(\left(x+3\right)\left(2y-1\right)=1\cdot5=5\cdot1=\left(-1\right)\cdot\left(-5\right)=\left(-5\right)\cdot\left(-1\right)\)
=>\(\left(x+3;2y-1\right)\in\left\{\left(1;5\right);\left(5;1\right);\left(-1;-5\right);\left(-5;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(-2;3\right);\left(2;1\right);\left(-4;-2\right);\left(-8;0\right)\right\}\)
a)x-2/5=3/8
(x-2).8=5.3
x-2=5.3:8
x-2=1
=>x=3
b,c )tương tự
\(\frac{5}{6}\div\frac{3}{8}\times\frac{3}{4}\)\(=\frac{5}{3}\)
\(\frac{5}{6}:\frac{3}{8}x\frac{3}{4}=\frac{5}{6}x\frac{8}{3}x\frac{3}{4}=\frac{5x8x3}{6x3x4}=\frac{5}{3}\)
\(\left(\frac{1}{4}+\frac{3}{8}\right)\times\frac{8}{7}\)
\(=\frac{5}{8}\times\frac{8}{7}\)
\(=\frac{5}{7}\)
#H
\(\left(\frac{1}{4}+\frac{3}{8}\right)\cdot\frac{8}{7}=\frac{5}{8}\cdot\frac{8}{7}=\frac{5}{7}\)
1) \(M=\frac{x^2+y^2+7}{x^2+y^2+5}=1+\frac{2}{x^2+y^2+5}\)
Ta có: \(x^2+y^2\ge0,\forall x;y\)
=> \(x^2+y^2+5\ge5\) với mọi x; y
=> \(\frac{2}{x^2+y^2+5}\le\frac{2}{5}\)
=> \(M\le1+\frac{2}{5}=\frac{7}{5}\)
Dấu "=" xảy ra <=> x = y = 0
Vậy max M = 7/5 đạt tại x = y = 0
2) \(f\left(x-1\right)=x^2-3x+5=x^2-x-2x+2+3\)
\(=x\left(x-1\right)-2\left(x-1\right)+3=x\left(x-1\right)-\left(x-1\right)-\left(x-1\right)+3\)
\(=\left(x-1\right)\left(x-1\right)-\left(x-1\right)+3\)
=> \(f\left(x\right)=x.x-x+3=x^2-x+3\)
a, x-2/5=3/8 suy ra x-2.8=3.5 tương đương x-2.8=15 tương đương x-2=1,875 suy ra x=3,875
a)\(\frac{x-2}{5}=\frac{3}{8}\)
\(\Leftrightarrow\)\(x-2=\frac{3}{8}\cdot5\)
\(\Leftrightarrow\)\(x-2=\frac{15}{8}\)
\(\Leftrightarrow x=\frac{31}{8}\)
Vậy x=\(\frac{31}{8}\)