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Tính
a) đáp án là 90
b)đáp án là âm 9
c)đáp án là âm 10
tìm x
a) xbằng 5
b) x bằng 3
c) x bằng 3
a; -2\(x\) - 3.(\(x-17\)) = 34 - 2.( - \(x\) + 25)
- 2\(x\) - 3\(x\) + 51 = 34 + 2\(x\) - 50
2\(x\) + 2\(x\) + 3\(x\) = - 34 + 50 + 51
7\(x\) = 67
\(x\) = 67 : 7
\(x\) = \(\dfrac{67}{7}\)
Vậy \(x\) = \(\dfrac{67}{7}\)
b; 17\(x\) + 3.(- 16\(x\) - 37) = 2\(x\) + 43 - 4\(x\)
17\(x\) - 48\(x\) - 111 = 2\(x\) - 4\(x\) + 43
- 31\(x\) - 2\(x\) + 4\(x\) = 111 + 43
- \(x\) x (31 + 2 - 4) = 154
- \(x\) x (33 - 4) = 154
- \(x\) x 29 = 154
- \(x\) = 154 : (-29)
\(x\) = - \(\dfrac{154}{29}\)
Vậy \(x=-\dfrac{154}{29}\)
a, A= 5x415 x 99 - 4x320x 89 = 5x230x318-22x 227x320
=229x318(2x5-32)=229x318
B=5x29x619-7x229x276=5x29x219x3197x229x318
=238x318(5x3-7x2)=228-318
->A:B= 229x318:228:318=2
tính :
a)\(\frac{459}{987}\cdot\left(\frac{1}{4}+\frac{1}{12}-\frac{1}{3}\right)-\frac{2009}{2010}\)
\(=\frac{459}{987}\cdot\left(\frac{3}{12}+\frac{1}{12}-\frac{4}{12}\right)-\frac{2009}{2010}\)
\(=\frac{459}{987}\cdot0-\frac{2009}{2010}\)
\(=\frac{-2009}{2010}\)
b) \(\frac{-9}{7}-\frac{5}{7}\cdot\left[\left(\frac{-2}{3}\right)^2-1\right]\div\frac{-5}{9}\)
\(=\frac{-9}{7}-\frac{5}{7}\cdot\left[\frac{4}{9}-1\right]\div\frac{-5}{9}\)
\(=\frac{-9}{7}-\frac{5}{7}\cdot\frac{-5}{9}\div\frac{-5}{9}\)
\(=\frac{-9}{7}-\frac{5}{7}\)
\(=-2\)
tìm x:
a) \(\left(x-\frac{7}{18}\right)-\frac{15}{27}=\frac{-10}{27}\)
\(\left(x-\frac{7}{18}\right)=\frac{-10}{27}+\frac{15}{27}\)
\(\left(x-\frac{7}{18}\right)=\frac{5}{27}\)
\(x=\frac{5}{27}+\frac{7}{18}\)
\(\Rightarrow x=\frac{31}{54}\)
b) \(\left(3\frac{1}{2}-2x\right)\cdot\frac{11}{3}=7\frac{1}{3}\)
\(\left(\frac{7}{2}-2x\right)\cdot\frac{11}{3}=\frac{22}{3}\)
\(\left(\frac{7}{2}-2x\right)=\frac{22}{3}\div\frac{11}{3}\)
\(\left(\frac{7}{2}-2x\right)=2\)
\(2x=\frac{7}{2}-2\)
\(x=\frac{3}{2}\div2=\frac{3}{4}\)
\(\Rightarrow x=\frac{3}{4}\)
Câu 1
a)(-9)+(-3) = -9 - 3 = -12
b)6.7.52+42.75 = 42. 25 + 42. 75 = 42(25 + 75) = 42. 100 = 4200
c)4529-(186+4529) = 4529 - 186 - 4529 = -186
Câu 2
a)x-7=15
<=> x = 15 + 7
<=> x = 22
b)9+4.(x-5)=17
<=> 9 + 4x - 20 = 17
<=> 4x = 28
<=> x = 7
c)(x-3)3=27
<=> (x - 3)3 = 33
=> x - 3 = 3
<=> x = 6
Câu 1:
a) Ta có: (-9)+(-3)
=-(9+3)
=-12
b) Ta có: \(6\cdot7\cdot5^2+42\cdot75\)
\(=6\cdot7\cdot5^2+6\cdot7\cdot5^2\cdot3\)
\(=6\cdot7\cdot5^2\cdot\left(1+3\right)\)
\(=6\cdot7\cdot25\cdot4\)
\(=42\cdot100=4200\)
c) Ta có: 4529-(186+4529)
=4529-186-4529
=-186
Câu 2:
a) Ta có: x-7=15
\(\Leftrightarrow x=15+7\)
hay x=22
Vậy: x=22
b) Ta có: \(9+4\left(x-5\right)=17\)
\(\Leftrightarrow9+4x-20=17\)
\(\Leftrightarrow4x-11=17\)
\(\Leftrightarrow4x=28\)
hay x=7
Vậy: x=7
c) Ta có: \(\left(x-3\right)^3=27\)
\(\Leftrightarrow\left(x-3\right)^3=3^3\)
\(\Leftrightarrow x-3=3\)
hay x=6
Vậy: x=6