a, [42-98] - [42-12] - 12
b, [-5] x 4x [-2] x3 x [-25]
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Bài 2:
a: =>x-35=-23
=>x=12
b: =>|x-8|=13
=>x-8=13 hoặc x-8=-13
=>x=21 hoặc x=-5
Bài 1:
a: =42-98-42+12-12=-98
b: =10x4x3x(-25)=40x(-25)x3=-1000x3=-3000
a, 2 x 31 x 12 + 4 x 6 x 42 + 8 x 27 x 3
= 2 x 12 x 31 + 4 x 6 x 42 + 8 x 3 x 27
= 24 x 31 + 24 x 42 + 24 x 27
= 24 x ( 31 + 24 + 27 )
= 24 x 82
= 1968
b, 2 x 53 x 12 + 4 x 6 x 87 - 3 x 8 x 40
= 24 x 53 + 24 x 87 - 24 x 40
= 24 x ( 53 + 87 - 40 )
= 24 x 100
= 2400
c, Tương tự
a) \(\left(x+13\right):5=12\)
\(\Rightarrow x+13=12\cdot5\)
\(\Rightarrow x+13=60\)
\(\Rightarrow x=60-13\)
\(\Rightarrow x=47\)
b) \(23\cdot\left(42-x\right)=23\)
\(\Rightarrow42-x=23:23\)
\(\Rightarrow42-x=1\)
\(\Rightarrow x=42-1\)
\(\Rightarrow x=41\)
1: \(=\dfrac{-\left[\left(x+5\right)^2-9\right]}{\left(x+2\right)^2}=\dfrac{-\left(x+5-3\right)\left(x+5+3\right)}{\left(x+2\right)^2}\)
\(=\dfrac{-\left(x+2\right)\left(x+8\right)}{\left(x+2\right)^2}=\dfrac{-\left(x+8\right)}{x+2}\)
2: \(=\dfrac{2x\left(x^2-4x+16\right)}{\left(x+4\right)\left(x^2-4x+16\right)}=\dfrac{2x}{x+4}\)
3: \(=\dfrac{5x\left(x^2+1\right)}{\left(x^2-1\right)\left(x^2+1\right)}=\dfrac{5x}{x^2-1}\)
4: \(=\dfrac{3\left(x^2-4x+4\right)}{x\left(x^3-8\right)}=\dfrac{3\left(x-2\right)^2}{x\left(x-2\right)\left(x^2+2x+4\right)}\)
\(=\dfrac{3\left(x-2\right)}{x\left(x^2+2x+4\right)}\)
5: \(=\dfrac{2a\left(a-b\right)}{a\left(c+d\right)-b\left(c+d\right)}=\dfrac{2a\left(a-b\right)}{\left(c+d\right)\left(a-b\right)}=\dfrac{2a}{c+d}\)
6: \(=\dfrac{x\left(x-y\right)}{\left(x-y\right)\left(x+y\right)}\cdot\left(-1\right)=\dfrac{-x}{x+y}\)
7: \(=\dfrac{2\left(1-a\right)}{-\left(1-a^3\right)}=\dfrac{-2\left(1-a\right)}{\left(1-a\right)\left(1+a+a^2\right)}=-\dfrac{2}{1+a+a^2}\)
8: \(=\dfrac{x^4\left(x^3-1\right)}{\left(x^3-1\right)\left(x^3+1\right)}=\dfrac{x^4}{x^3+1}\)
9: \(=\dfrac{\left(x+2-x+2\right)\left(x+2+x-2\right)}{16x}=\dfrac{4\cdot2x}{16x}=\dfrac{1}{2}\)
10: \(=\dfrac{0.5\left(49x^2-y^2\right)}{0.5x\left(7x-y\right)}=\dfrac{1}{x}\cdot\dfrac{\left(7x-y\right)\left(7x+y\right)}{7x-y}\)
\(=\dfrac{7x+y}{x}\)
a)Ta có:
\(\left\{{}\begin{matrix}\left|x+3\right|\ge0\\\left|x+9\right|\ge0\\\left|x+5\right|\ge0\end{matrix}\right.\)
\(\Rightarrow\left|x+3\right|+\left|x+9\right|+\left|x+5\right|\ge0\)
\(\Rightarrow4x\ge0\Rightarrow x\ge0\)
\(\Rightarrow\left|x+3\right|+\left|x+9\right|+\left|x+5\right|=x+3+x+9+x+5=3x+17=4x\)
\(\Rightarrow17=4x-3x\Rightarrow x=17\)
b)Tương tự câu a, ta chứng minh được \(x\ge0\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|+...+\left|x+98\right|+\left|x+99\right|=x+1+x+2+...+x+98+x+99=99x+4950=100x\)
\(\Rightarrow4950=100x-99x\Rightarrow x=4950\)
c)Ta có:
\(\left|x-1\right|+\left|x-5\right|=\left|x-1\right|+\left|5-x\right|=4=\left|x-1+5-x\right|\)
\(\Rightarrow1\le x\le5\Rightarrow x\in\left\{1;2;3;4;5\right\}\)
a) Tính bằng cách thuận tiện nhất:
134 x 4 x 5
= 134 x (4 x 5)
= 134 x 20 = 1680
5 x 36 x 2
= 36x (5 x 2)
= 36 x 10 = 360
42 x 2 x 7 x5
= (42 x 7) x (2 x5)
= 294 x 10 = 2940
b) 137 x 3 + 137 x 97
= 137 x (3 + 97)
= 137 x 100 = 13700
94 x 12 + 94 x 88
= 94 x (12 + 88)
= 94 x 100 = 9400
428 x 12 - 428 x 2
= 428 x (12 - 2) = 4280
537 x 39 - 537 x 19
= 537 x (39 - 19)
= 537 x 20 = 10740
\(a,8x-3=5x+12\\ \Leftrightarrow8x-5x=12+3\\ \Leftrightarrow3x=15\\ \Leftrightarrow x=\dfrac{15}{3}=5\)
\(b,x-12+4x=25+2x-1\\ \Leftrightarrow x+4x-2x=25-1+12\\ \Leftrightarrow3x=36\\ \Leftrightarrow x=\dfrac{36}{3}=12\)
\(c,7-\left(2x+4\right)=-\left(x+4\right)\\ \Leftrightarrow7-2x-4=-x-4\\ \Leftrightarrow-2x+x=-4+4-7\\ \Leftrightarrow-x=-7\\ \Leftrightarrow x=7\)
\(d,3-4x\left(45-2x\right)=8x^2+x-300\\ \Leftrightarrow3-100x+8x^2=8x^2+x-300\\ \Leftrightarrow8x^2-8x^2-100x-x=-300-3\\ \Leftrightarrow-101x=-303\\ \Leftrightarrow x=\dfrac{-303}{-101}=3\)
Đề câu d của bạn hình như sai dấu ý
a,=-56-30-12 b,=-20.-2.3.25
=-86-12 =40.3.25
=-98 = 120.25
=3000