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5 + ( x + 27 ) = 64
( x + 27 ) = 64 - 5 ( x + 27 ) = 59 x = 59 - 27 x = 32a) \(\Rightarrow x+27=59\Rightarrow x=32\)
b) \(\Rightarrow x-2=39\Rightarrow x=41\)
c) \(\Rightarrow x+5=-322\Rightarrow x=-327\)
d) \(\Rightarrow5x=35\Rightarrow x=7\)
e) \(\Rightarrow4\left(x-5\right)=56\Rightarrow x-5=14\Rightarrow x=19\)
f) \(\Rightarrow15+x=37\Rightarrow x=22\)
g) \(\Rightarrow7\left(13-x\right)=35\Rightarrow13-x=5\Rightarrow x=8\)
h) \(\Rightarrow10\left(x+1\right)=100\Rightarrow x+1=10\Rightarrow x=9\)
(x+1)+(x+4)+...+(x+28)=155
<=>x+1+x+4+...+x+28=155
<=>10x+(1+4+...+28)=155
<=>10x+(((1+28)(28-1):3+1))/2=155
<=>10x+145=155
<=>10x=10
<=>x=1
Vậy x=1
(x+1) + (x+4) + ( x+7 ) + (x+10)+...+(x+28)=155
x+1 + x+4 + x+7 + x+10 + ...+ x+28 = 155
(x+x+x+x+...+x) + (1+4+7+10+...+28) = 155
((28-1):3+1=10 (x) + (28 +1 ) . 10 : 2 = 145 )*cách tính*
x.10 + 145 = 155
x.10 = 155 - 145
x.10 =10
x = 10:10
x =1
(X + 1) + (X + 4) + (X + 7) + ... + (X+ 28) = 155
* Nhận xét: Dãy số 1;4;7;...; 28 có (28-1) :3 +1 = 10 (số hạng)
<=> 10X + (1+4+7+...+28) = 155 (10X ở tiểu học được viết 10 x X)
Đặt B = 1 + 4 + 7 +... + 28 => B = (1 + 28) + (4 + 25) +...
B có 10 số hạng được ghép 5 cặp mỗi cặp có tổng bằng 29. Vậy ta có:
<=> 10X + 29x5 =155
<=> 10X = 10
<=> X = 1
**** nhé
b) \(\left(x+1\right)+\left(x+4\right)+...+\left(x+28\right)=155\)
\(\Leftrightarrow10x+\left(1+4+...+28\right)=155\)
\(\Leftrightarrow10x+145=155\)
\(\Leftrightarrow10x=10\)
\(\Leftrightarrow x=1\)
Vậy x=1
\(\frac{x}{1\cdot4}+\frac{x}{4\cdot7}+\frac{x}{7\cdot10}+\frac{x}{10\cdot13}+\frac{x}{13\cdot16}=\frac{5}{2}\)
\(\Rightarrow\frac{x}{3}\left[\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+\frac{3}{13\cdot16}\right]=\frac{5}{2}\)
\(\Rightarrow\frac{x}{3}\left[1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{16}\right]=\frac{5}{2}\)
\(\Rightarrow\frac{x}{3}\left[1-\frac{1}{16}\right]=\frac{5}{2}\)
\(\Rightarrow\frac{x}{3}\cdot\frac{15}{16}=\frac{5}{2}\)
\(\Rightarrow\frac{x}{3}=\frac{5}{2}:\frac{15}{16}\)
\(\Rightarrow\frac{x}{3}=\frac{5}{2}\cdot\frac{16}{15}\)
\(\Rightarrow\frac{x}{3}=\frac{1}{1}\cdot\frac{8}{3}\)
\(\Rightarrow\frac{x}{3}=\frac{8}{3}\Leftrightarrow x=8\)
1: =-2/9(15/17+2/17)=-2/9
2: \(=\dfrac{-6}{3}+\dfrac{-21}{90}\)
=-2-7/30=-67/30
3: \(=\dfrac{3}{4}\cdot\dfrac{7}{5}+\dfrac{9}{7}\cdot\dfrac{3}{2}\)
=21/20+27/14=417/140
4: =-25/13(5/19+14/19)=-25/13
5: =-7/5-45/21=-7/5-15/7=-124/35
1: =-2/9(15/17+2/17)=-2/9
2: =−63+−2190=−63+−2190
=-2-7/30=-67/30
3: =34⋅75+97⋅32=34⋅75+97⋅32
=21/20+27/14=417/140
4: =-25/13(5/19+14/19)=-25/13
5: =-7/5-45/21=-7/5-15/7=-124/35
a, \(\left(2x+7\right)^4=10^{11}:10^7\)
\(\Rightarrow\left(2x+7\right)^4=10^4\)
\(\Rightarrow2x+7=10\)
\(\Rightarrow2x=10-7\)
\(\Rightarrow2x=3\)
\(\Rightarrow x=\dfrac{3}{2}\) hay \(x=1,5\)
b, \(5^{x-1}.7^{x-1}=25.49\)
\(\Rightarrow\)\(5^{x-1}.7^{x-1}=5^2.7^2\)
\(\Rightarrow\left\{{}\begin{matrix}5^{x-1}=5^2\\7^{x-1}=7^2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x-1=2\\x-1=2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=3\\x=3\end{matrix}\right.\)
c, \(\left(x-5\right)^{2018}=9.\left(x-5\right)^{2016}\)
\(\Rightarrow\dfrac{\left(x-5\right)^{2018}}{\left(x-5\right)^{2016}}=9.\dfrac{\left(x-5\right)^{2016}}{\left(x-5\right)^{2016}}\)
\(\Rightarrow\left(x-5\right)^2=9\)
\(\Leftrightarrow\left(x-5\right)^2=3^2\)
\(\Rightarrow x-5=3\)
\(\Rightarrow x=3+5\)
\(\Rightarrow x=8\)
1: 7-x=8+(-7)
=>7-x=8-7=1
=>x=7-1=6
2: \(x-8=\left(-3\right)-8\)
=>x-8=-11
=>\(x=-11+8=-3\)
3: \(2-x=10-9+23\)
=>\(2-x=33-9=24\)
=>x=2-24=-22
4: \(-2-x=15\)
=>\(x=-2-15=-17\)
5: \(-7+x-8=-3-1+13\)
=>x-14=13-4=9
=>x=9+14=23
6: 100-x+7=-x+3
=>107-x=3-x
=>107=3(vô lý)
7: \(23+x=8-2x\)
=>\(x+2x=8-23\)
=>3x=-15
=>x=-15/3=-5
x(x+2)=0
suy ra x=0 hoặc x+2=0
5-2x=-7
2x=-7+5
2x=-(7-5)
2x=-2
x=-2:2
x=-1
Vậy x=-1
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Jim Rohn – Triết lý cuộc đời
(x+1)+(x+4)+(x+7)+(x+10)+(x+13)=155
(x+x+x+x+...+x)+(1+4+7+10+...+13)=155
5x + 35 =155
5x =155 - 35
5x =120
x =120 : 5
x =24
Vậy ...
5x+35=155
5x=120
x=40