a, 11.(x-9)=77
b, 2x+2=250/8
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
13 tháng 8 2018
a) ( 72 - 8x9 ) : ( 20 + 21 + 22 + 23 + 24 + 25 )
= ( 72 - 72 ) : ( 20 + 21 + 22 + 23 + 24 + 25 )
= 0 : ( 20 + 21 + 22 + 23 + 24 + 25 )
= 0
b) ( 500 x 9 - 250 x 18 ) x ( 1 + 2 + 3 + ... + 9 )
= ( 250 x 2 x 9 - 250 x 18 ) x ( 1 + 2 + 3 + ... + 9 )
= ( 250 x 18 - 250 x 18 ) x ( 1 + 2 + 3 + ... + 9 )
= 0 x ( 1 + 2 + 3 + ... + 9 )
= 0
c ) ( 11 + 13 + 15 + ... + 19 ) x ( 6 x 8 - 48 )
= ( 11 + 13 + 15 + ... + 19 ) x ( 48 - 48 )
= ( 11 + 13 + 15 + ... + 19 ) x 0
= 0
13 tháng 8 2018
a) ( 72 - 8x9 ) : ( 20 + 21 + 22 + 23 + 24 + 25 )
= ( 72 - 72 ) : ( 20 + 21 + 22 + 23 + 24 + 25 )
= 0 : ( 20 + 21 + 22 + 23 + 24 + 25 )
= 0
6 tháng 4 2021
a) ( 72 - 8 x 9 ) : ( 20 + 21 + 22 + 23 + 24 + 25 )
= ( 72 - 72 ) : ( 20 + 21 + 22 + 23 + 24 + 25 )
= 0 : ( 20 + 21 + 22 + 23 + 24 + 25 )
= 0
b) ( 500 x 9 - 250 x 18 ) x ( 1 + 2 + 3 + ... + 9 )
= ( 500 x 9 - 250 x 2 x 9 ) x ( 1 + 2 + 3 + ... + 9 )
= ( 500 x 9 - 500 x 9 ) x ( 1 + 2 + 3 + ... + 9 )
= 0 x ( 1 + 2 + 3 + ... + 9 )
= 0
28 tháng 10 2023
a: \(5^{\left(x-2\right)\left(x+3\right)}=1\)
=>\(\left(x-2\right)\left(x+3\right)=0\)
=>\(\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
c: \(\left|x^2+2x\right|+\left|y^2-9\right|=0\)
mà \(\left\{{}\begin{matrix}\left|x^2+2x\right|>=0\forall x\\\left|y^2-9\right|>=0\forall y\end{matrix}\right.\)
nên \(\left\{{}\begin{matrix}x^2+2x=0\\y^2-9=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\left(x+2\right)=0\\\left(y-3\right)\left(y+3\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x\in\left\{0;-2\right\}\\y\in\left\{3;-3\right\}\end{matrix}\right.\)
d: \(2^x+2^{x+1}+2^{x+2}+2^{x+3}=120\)
=>\(2^x\left(1+2+2^2+2^3\right)=120\)
=>\(2^x\cdot15=120\)
=>\(2^x=8\)
=>x=3
e: \(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)
=>\(\left(x-7\right)^{x+11}-\left(x-7\right)^{x+1}=0\)
=>\(\left(x-7\right)^{x+1}\left[\left(x-7\right)^{10}-1\right]=0\)
=>\(\left[{}\begin{matrix}x-7=0\\x-7=1\\x-7=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=8\\x=6\end{matrix}\right.\)
AD
6 tháng 11 2023
1.Tính nhanh nếu có thể:
a) 22 + 23 + 89 + 77
= ( 77 + 23 ) + 22 + 89
= 100 + 22 + 89
= 122 + 89
= 211
b) 35 . 15 + 15 . 65
= 15 . ( 35 + 65 )
= 15 . 100
= 1500
c) 7^2 - 36 : 3^2
= 7^2 - 36 : 9
= 7^2 - 4
= 49 - 4
= 45
d) 476 - {5 . [409 - (8 . 3 - 21)2] - 1724}
= 476 - {5 . [409 - (24 - 21)^2] - 1724}
= 476 - {5 . [409 - (3^2)] - 1724}
= 476 - {5 . [409 - 9 ] - 1724}
= 476 - {5. 400 - 1724}
= 476 - {2000 - 1724}
= 476 - 276
= 200
1 tháng 9 2020
a) x - 8 - (12 - 2x) = -20
=> x - 8 - 12 + 2x = -20
=> (x + 2x) + (-8 - 12) = -20
=> 3x - 20 = -20
=> 3x = 0 => x = 0
b) -27 + (x + 8) - ( +11) = 2
=> -27 + x + 8 - 11 = 2
=> -27 + x = 2 + 11 - 8
=> -27 + x = 5
=> x = 5 - (-27) = 32
c) -2x - 16 = -2 - (3x + 9)
=> -2x - 16 = -2 - 3x - 9
=> -2x - 16 + 2 + 3x + 9 = 0
=> (-2x + 3x) + (-16 + 2 + 9) = 0
=> x - 5 = 0
=> x = 5
1 tháng 9 2020
\(a,x-8-\left(12-2x\right)=-20\)
\(x-8-12+2x=-20\)
\(x+2x-8-12=-20\)
\(3x-20=-20\)
\(3x=-20+20\)
\(3x=0\)
\(x=0\)
\(b,-27+\left(x+8\right)-\left(+11\right)=2\)
\(-27+x+8-11=2\)
\(x-27+8-11=2\)
\(x-30=2\)
\(x=2+30\)
\(x=32\)
\(c,-2x-16=2-\left(3x+9\right)\)
\(-2x-16=2-3x-9\)
\(-2x+3x=2-9+16\)
\(x=9\)
Học tốt
1 tháng 4 2020
a) Đk: x \(\ne\)-2
Ta có: \(\frac{2}{x+2}-\frac{2x^2+16}{x^2+8}=\frac{5}{x^2-2x+4}\)
<=> \(\frac{2\left(x^2-2x+4\right)-\left(2x^2+16\right)}{\left(x+2\right)\left(x^2-2x+4\right)}=\frac{5\left(x+2\right)}{\left(x+2\right)\left(x^2-2x+4\right)}\)
<=> 2x2 - 4x + 8 - 2x2 - 16 = 5x + 10
<=> -4x - 8 = 5x + 10
<=> -4x - 5x = 10 + 8
<=> -9x = 18
<=> x = -2 (ktm)
=> pt vô nghiệm
b) Đk: x \(\ne\)2; x \(\ne\)-3
Ta có: \(\frac{1}{x-2}-\frac{6}{x+3}=\frac{5}{6-x^2-x}\)
<=> \(\frac{x+3}{\left(x-2\right)\left(x+3\right)}-\frac{6\left(x-2\right)}{\left(x-2\right)\left(x+3\right)}=-\frac{5}{\left(x-2\right)\left(x+3\right)}\)
<=> x + 3 - 6x + 12 = -5
<=> -5x = -5 - 15
<=> -5x = -20
<=> x = 4
vậy S = {4}
c) Đk: x \(\ne\)8; x \(\ne\)9; x \(\ne\)10; x \(\ne\)11
Ta có: \(\frac{8}{x-8}+\frac{11}{x-11}=\frac{9}{x-9}+\frac{10}{x-10}\)
<=> \(\left(\frac{8}{x-8}+1\right)+\left(\frac{11}{x-11}+1\right)=\left(\frac{9}{x-9}+1\right)+\left(\frac{10}{x-10}+1\right)\)
<=> \(\frac{x}{x-8}+\frac{x}{x-11}-\frac{x}{x-9}-\frac{x}{x-10}=0\)
<=> \(x\left(\frac{1}{x-8}+\frac{1}{x-11}-\frac{1}{x-9}-\frac{1}{x-10}\right)=0\)
<=> x = 0 (vì \(\frac{1}{x-8}+\frac{1}{x-11}-\frac{1}{x-9}-\frac{1}{x-10}\ne0\)
Vậy S = {0}
24 tháng 9 2021
1) \(ĐK:x\in R\)
2) \(ĐK:x< 0\)
3) \(ĐK:x\in\varnothing\)
4) \(=\sqrt{\left(x+1\right)^2+2}\)
\(ĐK:x\in R\)
5) \(=\sqrt{-\left(a-4\right)^2}\)
\(ĐK:x\in\varnothing\)
a, => x - 9 = 77 : 11
=> x - 9 = 7
=> x = 7 + 9
=> x = 16