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Câu 4:
a) Ta có: \(\left|-x+8\right|\ge0\)
\(\Rightarrow A=\left|-x+8\right|-21\ge-21\)
Vậy \(MIN_A=-21\) khi x = 8
b) Ta có: \(\left|-x-17\right|+\left|y-36\right|\ge0\)
\(\Rightarrow B=\left|-x-17\right|+\left|y-36\right|+12\ge12\)
Vậy \(MIN_B=12\) khi \(x=-17;y=36\)
c) Ta có: \(-\left|2x-8\right|\le0\)
\(\Rightarrow C=-\left|2x-8\right|-35\le-35\)
Vậy \(MAX_C=-35\) khi \(x=4\)
d) Ta có: \(3\left(3x-12\right)^2\ge0\)
\(\Rightarrow D=3\left(3x-12\right)^2-37\ge-37\)
Vậy \(MIN_D=-37\) khi x = 4
e) Ta có: \(-3\left|2x+50\right|\le0\)
\(\Rightarrow E=-21-3\left|2x+50\right|\le-21\)
Vậy \(MAX_E=-21\) khi x = -25
g) \(\left(x-3\right)^2+\left|x^2-9\right|\ge0\)
\(\Rightarrow G=\left(x-3\right)^2+\left|x^2-9\right|+25\ge25\)
Vậy \(MIN_G=25\) khi x = 3
Câu hỏi của Đào Na - Toán lớp 6 - Học toán với OnlineMath
\(\frac{-5}{6}+\frac{8}{3}+\frac{29}{-6}\le x\le\frac{-1}{2}+2+\frac{5}{2}\)
\(\Rightarrow-3\le x\le4\)
\(\Rightarrow x\in\left\{-3;-2;-1;0;1;2;3;4\right\}\)
\(\left(\frac{-2}{3}-\frac{1}{2}\right):\frac{-1}{4}\le x\le\left(\frac{-5}{6}+\frac{9}{4}:\frac{-3}{2}\right)\cdot\frac{-13}{2}\)
\(\Rightarrow\frac{14}{3}\le x\le\frac{91}{6}\)
\(\Rightarrow\frac{28}{6}\le x\le\frac{91}{6}\)
\(\Rightarrow x\in\left\{\frac{28}{6};\frac{29}{6};...;\frac{90}{6};\frac{91}{6}\right\}\)
a: x/2=-5/y
=>xy=-10
=>\(\left(x,y\right)\in\left\{\left(1;-10\right);\left(-10;1\right);\left(-1;10\right);\left(10;-1\right);\left(2;-5\right);\left(-5;2\right);\left(-2;5\right);\left(5;-2\right)\right\}\)
b: =>xy=12
mà x>y>0
nên \(\left(x,y\right)\in\left\{\left(12;1\right);\left(6;2\right);\left(4;3\right)\right\}\)
c: =>(x-1)(y+1)=3
=>\(\left(x-1;y+1\right)\in\left\{\left(1;3\right);\left(3;1\right);\left(-1;-3\right);\left(-3;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(2;2\right);\left(4;0\right);\left(0;-4\right);\left(-2;-2\right)\right\}\)
d: =>y(x+2)=5
=>\(\left(x+2;y\right)\in\left\{\left(1;5\right);\left(5;1\right);\left(-1;-5\right);\left(-5;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(-1;5\right);\left(3;1\right);\left(-3;-5\right);\left(-7;-1\right)\right\}\)