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2: B=|x+5|-|x-2|<=|x+5-x+2|=7
Dấu = xảy ra khi -5<=x<=2
\(=\dfrac{1}{2}\cdot\dfrac{6}{5}\cdot\dfrac{8}{7}+\dfrac{3}{7}\cdot\dfrac{9}{5}=\dfrac{24}{35}+\dfrac{27}{35}=\dfrac{51}{35}\)
Bài làm
\(7+\left(\frac{7}{12}-\frac{1}{2}+3\right)-\left(\frac{1}{12}+5\right)\)
\(=\frac{84}{12}+\left(\frac{7}{12}-\frac{6}{12}+\frac{36}{12}\right)-\left(\frac{1}{12}+\frac{60}{12}\right)\)
\(=\frac{84}{12}+\frac{37}{12}-\frac{61}{12}\)
\(=\frac{60}{12}\)
\(=5\)
# Chúc bạn học tốt #
a)
\(\begin{array}{l}A = \left( {2 + \frac{1}{3} - \frac{2}{5}} \right) - \left( {7 - \frac{3}{5} - \frac{4}{3}} \right) - \left( {\frac{1}{5} + \frac{5}{3} - 4} \right).\\A = \left( {\frac{{30}}{{15}} + \frac{5}{{15}} - \frac{6}{{15}}} \right) - \left( {\frac{{105}}{{15}} - \frac{9}{{15}} - \frac{{20}}{{15}}} \right) - \left( {\frac{3}{{15}} + \frac{{25}}{{15}} - \frac{{60}}{{15}}} \right)\\A = \frac{{29}}{{15}} - \frac{{76}}{{15}} - \left( {\frac{{ - 32}}{{15}}} \right)\\A = \frac{{29}}{{15}} - \frac{{76}}{{15}} + \frac{{32}}{{15}}\\A = \frac{{ - 15}}{{15}}\\A = - 1\end{array}\)
b)
\(\begin{array}{l}A = \left( {2 + \frac{1}{3} - \frac{2}{5}} \right) - \left( {7 - \frac{3}{5} - \frac{4}{3}} \right) - \left( {\frac{1}{5} + \frac{5}{3} - 4} \right)\\A = 2 + \frac{1}{3} - \frac{2}{5} - 7 + \frac{3}{5} + \frac{4}{3} - \frac{1}{5} - \frac{5}{3} + 4\\A = \left( {2 - 7 + 4} \right) + \left( {\frac{1}{3} + \frac{4}{3} - \frac{5}{3}} \right) + \left( { - \frac{2}{5} + \frac{3}{5} - \frac{1}{5}} \right)\\A = - 1 + 0 + 0 = - 1\end{array}\)
\(\frac{3}{5}\cdot16\frac{5}{7}-\frac{3}{5}\cdot26\frac{5}{7}\)
\(\Rightarrow=\frac{3}{5}\cdot\left(16\frac{5}{7}-26\frac{5}{7}\right)\)
\(\Rightarrow=\frac{3}{5}\cdot\left(\frac{117}{7}-\frac{187}{7}\right)\)\(=\frac{3}{5}\cdot\left(-\frac{70}{7}\right)\)\(=\frac{3}{5}\cdot\left(-10\right)=-\frac{30}{5}=-6\)
\(\frac{3}{5}\cdot16\frac{5}{7}-\frac{3}{5}\cdot26\frac{5}{7}\)
\(=\frac{3}{5}\cdot\left(16\frac{5}{7}-26\frac{5}{7}\right)\)
\(=\frac{3}{5}\cdot\left(-10\right)\)
\(=\frac{-30}{5}=\frac{-6}{5}\)
ta có
\(A=\left|x-8\right|+\left|x+2\right|+\left|x+5\right|+\left|x+7\right|\ge\left|-x+8-x-2+x+5+x+7\right|=18\)
Dấu bằng xảy ra khi \(-5\le x\le-2\)
\(B=\left|x+3\right|+\left|x-5\right|+\left|x-2\right|\ge\left|x+3-x+5\right|+\left|x-2\right|=8+\left|x-2\right|\ge8\)
Dấu bằng xảy ra khi \(x=2\)
\(C=\left|x+5\right|-\left|x-2\right|\le\left|x+5+2-x\right|=7\)
Dấu bằng xảy ra khi \(x\ge2\)
bài 1 :
B=15-3x-3y
a) x+y-5=0
=>x+y=-5
B=15-3x-3y <=> B=15-3(x+y)
Thay x+y=-5 vào biểu thức B ta được :
B=15-3(-5)
B=15+15
B=30
Vậy giá trị của biểu thức B=15-3x-3y tại x+y+5=0 là 30
b)Theo đề bài ; ta có :
B=15-3x-3.2=10
15-3x-6=10
15-3x=16
3x=-1
\(x=\frac{-1}{3}\)
Bài 2:
a)3x2-7=5
3x2=12
x2=4
x=\(\pm2\)
b)3x-2x2=0
=> 3x=2x2
=>\(\frac{3x}{x^2}=2\)
=>\(\frac{x}{x^2}=\frac{2}{3}\)
=>\(\frac{1}{x}=\frac{2}{3}\)
=>\(3=2x\)
=>\(\frac{3}{2}=x\)
c) 8x2 + 10x + 3 = 0
=>\(8x^2-2x+12x-3=0\)
\(\Rightarrow\left(2x+3\right)\left(4x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x+3=0\\4x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=-3\\4x=1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{-3}{2}\\x=\frac{1}{4}\end{cases}}}\)
vậy \(x\in\left\{-\frac{3}{2};\frac{1}{4}\right\}\)
Bài 5 đề sai vì |1| không thể =2
Giải :
\(A=\left(6-\frac{2}{3}+\frac{1}{2}\right)-\left(5+\frac{5}{3}-\frac{3}{2}\right)-\left(3-\frac{7}{3}+\frac{5}{2}\right)\)
Ta có :
A= 6 - 5 - 3 - \(\frac{2}{3}\) - \(\frac{5}{3}\) + \(\frac{7}{3}\) + \(\frac{1}{2}\) + \(\frac{3}{2}\) - \(\frac{5}{2}\)
= - 2 - \(\frac{1}{2}\) = \(-\frac{5}{2}\)
57. \(\dfrac{1}{5^5}\) = 52 = 25
5⁷ . 1/5⁵
= 5⁷/5⁵
= 5²
= 25