\(A=\dfrac{2\left|y+7\right|+13}{2\left|y+7\right|+6}\le\dfrac{13}{6}\)
Dấu "=" xảy ra khi:
\(y=-7\)
\(B=\dfrac{15\left|x+1\right|+32}{6\left|x+1\right|+8}\le\dfrac{32}{8}=4\)
Dấu "=" xảy ra khi: \(x=-1\)
Câu 1: Tinh
\(A=\left(\dfrac{3}{5}\right)^2.5^2-\left(2\dfrac{1}{4}\right)^3:\left(\dfrac{3}{4}\right)^3+\dfrac{1}{2}\)
\(B=\left[\dfrac{4}{11}.\left(\dfrac{1}{25}\right)^0+\dfrac{7}{22}.2\right]^{2010}-\left(\dfrac{1}{2^2}:\dfrac{8^2}{4^4}\right)^{2009}\)
\(\dfrac{7}{8}.\left(\dfrac{2}{12}+\dfrac{4}{10}\right)\)
\(\dfrac{3}{2}-\dfrac{5}{6}:\left(\dfrac{1}{2}\right)^2+\sqrt{4}\)
Câu 2: Tim x
\(2.x-\dfrac{5}{4}=\dfrac{20}{15}\)
\(\left(x+\dfrac{1}{3}\right)^3=\left(\dfrac{-1}{8}\right)\)
Tìm x , biết :
1, |\(\dfrac{3}{2}\)x + \(\dfrac{1}{2}\)| = | 4x - 1 |
2, | \(\dfrac{5}{4}\)x - \(\dfrac{7}{2}\) | - | \(\dfrac{5}{8}x+\dfrac{3}{5}\) | = 0
3, | \(\dfrac{7}{5}x+\dfrac{2}{3}\) | = | \(\dfrac{4}{3}x-\dfrac{1}{4}\) |
4, | \(\dfrac{7}{8}x+\dfrac{5}{6}\) | - | \(\dfrac{1}{2}x+5\) | = 0
tính GTBT A=\(\dfrac{\text{15│x+1│+32}}{\text{6│x+1│+8}}\)
Tìm X, biết
a, X+ \(\dfrac{5}{6}=\dfrac{2}{5}-\dfrac{-2}{3}\)x
b,\(1\dfrac{2}{5}x+\dfrac{3}{7}=\dfrac{-4}{5}-\dfrac{1}{5}x\)
c,\(\dfrac{1}{3}-\dfrac{2}{3}\left(x-\dfrac{1}{2}\right)=1\)
Tìm giá trị nhỏ nhất của biểu thức:
a) \(A=3.\left|2x-\dfrac{3}{2}\right|+2021^0\)
b) \(B=2.\left|x-6\right|+3.\left(2y-1\right)^2+2021^0\)
Giúp mk nốt bài này nha
1, \(\dfrac{x}{3}=\dfrac{y}{2}\); \(\dfrac{x}{5}=\dfrac{x}{7}\) và x +2y = -112
2, 2x = 3y; 5y = 7z và 3x - 7y + 5z = 30
3, \(\dfrac{x}{y}=\dfrac{10}{9};\) \(\dfrac{y}{z}=\dfrac{3}{4}\) và x +2y - 3z = -48
Làm giúp, cần gấp
|x+\(\dfrac{1}{3}\)|-\(\dfrac{1}{2}\)x=\(\dfrac{7}{6}\)
1,tìm x:
a,\(\dfrac{3}{4}-|2x+1|=\dfrac{7}{8}\)
b,\(2.|2x-3|=\dfrac{1}{2}\)
c,\(7,5-3.|5-2x|=-4,5\)
d,\(|x+\dfrac{4}{15}|-|-3,75|=-|-2,15|\)
giúp mình nha mọi người
1/ Tìm GTNN:
a,\(\left|7x-5y\right|+\left|2z-3y\right|+\left|xy+yz+xz-2000\right|\)
b,\(\left|3x-7\right|+\left|3x+2\right|+8\)
2/ Tìm x, y:
a, \(\left|x-5\right|+\left|1-x\right|=\dfrac{12}{\left|y+1\right|+3}\)
b,\(\left|y+3\right|+5=\dfrac{10}{\left(2x-6\right)^2+2}\)
c,\(\left|x-1\right|+\left|3-x\right|=\dfrac{6}{\left|y+3\right|+3}\)
