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b, (y+1) + (y+2) + (y+3) + .... + (y+99) = 6138
(y+y+y+y+....+y) + (1+2+3+...+99) = 6138
99.y + 4950 = 6138
99.y = 6138 - 4950
99.y = 1188
y = 1188 : 99
x = 12
a) \(\left(y+1\right)+\left(y+4\right)+\left(y+7\right)+....+\left(y+28\right)=155155\)
\(\Rightarrow\left(y+y+...+y\right)+\left(1+4+7+...+28\right)=155155\)
\(\Rightarrow10x+145=155155\)
\(\Rightarrow10x=155010\)
\(\Rightarrow x=15501\)
Vậy x = 15501
b) \(\left(y+1\right)+\left(y+2\right)+..+\left(y+99\right)=6138\)
\(\Rightarrow\left(y+y+...+y\right)+\left(1+2+3+...+99\right)=6138\)
\(\Rightarrow99x+4950=6138\)
\(\Rightarrow99x=1188\)
\(\Rightarrow x=12\)
Vậy x = 12
Gợi ý: Các biểu thức mũ chẵn đều không âm.
\(a^{2n}+b^{2n}\le0\Leftrightarrow a^{2n}+b^{2n}=0\Leftrightarrow a=b=0\)
a,\(\left(x-\frac{2}{5}\right)^{2010}+\left(y+\frac{3}{7}\right)^{468}\)< \(0\)
Vì \(\left(x-\frac{2}{5}\right)^{2010}\);\(\left(y+\frac{3}{7}\right)^{468}\)đều > \(0\)
=> \(\left(x-\frac{2}{5}\right)^{2010}=0\)
\(\left(y+\frac{3}{7}\right)^{468}=0\)
=> \(\left(x-\frac{2}{5}\right)^{2010}=0^{2010}\)
\(\left(y+\frac{3}{7}\right)^{468}=0^{468}\)
=> \(x-\frac{2}{5}=0\)
\(y-\frac{3}{7}=0\)
=> \(x=\frac{2}{5}\)
\(y=\frac{3}{7}\)
Vậy \(x=\frac{2}{5}\)\(y=\frac{3}{7}\)
\(\left(y-\frac{1}{2}\right):\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)=\frac{1}{3}\)
\(\Leftrightarrow\left(y-\frac{1}{2}\right):\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{9}-\frac{1}{10}\right)=\frac{1}{3}\)
\(\Leftrightarrow\left(y-\frac{1}{3}\right):\left(1-\frac{1}{10}\right)=\frac{1}{3}\)
\(\Leftrightarrow\left(y-\frac{1}{2}\right):\frac{9}{10}=\frac{1}{3}\)
\(\Leftrightarrow\left(y-\frac{1}{2}\right)=\frac{3}{10}\)
\(\Leftrightarrow y=\frac{4}{5}\)
a) \(\left(2x+1\right)^3=125\)
\(\Rightarrow\left(2x+1\right)^3=5^3\)
\(\Rightarrow2x+1=5\)
\(\Rightarrow2x=5-1\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=4:2\)
\(\Rightarrow x=2\)
Vậy x = 2
b) \(\left(x-5\right)^4=\left(x-5\right)^6\)
\(\Rightarrow\left(x-5\right)^4-\left(x-5\right)^6=0\)
\(\Rightarrow\left(x-5\right)^4\left[1-\left(x-5\right)^2\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(x-5\right)^4=0\\1-\left(x-5\right)^2=0\end{cases}}\Rightarrow\orbr{\begin{cases}\left(x-5\right)^4=0\\\left(x-5\right)^2=1\end{cases}}\)
TH 1 : \(\left(x-5\right)^4=0\Rightarrow x-5=0\Rightarrow x=5\)
TH 2 : \(\left(x-5\right)^2=1\Rightarrow\orbr{\begin{cases}x-5=1\\x-5=-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=6\\x=4\end{cases}}\)
Vậy \(x\in\left\{5;6;4\right\}\)
c) \(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\Rightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)
\(\Rightarrow\left(2x-15\right)^3\left[\left(2x-15\right)^2-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}\left(2x-15\right)^3=0\\\left(2x-15\right)^2=1\end{cases}}\)
TH 1 : \(\left(2x-15\right)^3=0\Rightarrow2x-15=0\Rightarrow2x=15\Rightarrow x=\frac{15}{2}\)
TH 2 : \(\left(2x-15\right)^2=1\Rightarrow\orbr{\begin{cases}2x-15=1\\2x-15=-1\end{cases}}\Rightarrow\orbr{\begin{cases}2x=16\\2x=14\end{cases}}\Rightarrow\orbr{\begin{cases}x=8\\x=7\end{cases}}\)
Vậy \(x\in\left\{\frac{15}{2};8;7\right\}\)
_Chúc bạn học tốt_
35.(24−6.x)=0
24-6x= 0:35
24-6x=0
6x=24-0
6x=24
x=24:6=4
3.(x+7)−15=27
3(x+7)=27+15
3(x+7)=42
x+7=42:3
x+7=14
x=14-7=7
TK MIK NHA~~
b) x - ( 54 +31 - 69x -118) = 654+ (-3x)
x - 54 - 31 + 69x + 118 = 654 - 3x
x + 33 + 69x = 654 - 3x
70x + 3x = 654 - 33
73x = 621
c) 3 /4x+1 = 12/-24
3 /4x+1 = -1/2
=> 3.2 = -4x - 1
4x+1 = -6
4x = -7
x= -7/4
a) -(-313+65)x + (-(-259)) = (375-738)x
248x + 259 = -381x
248x + 381x = -259
629x = -259
x= -7/17
1, \(\frac{1}{2}-\left(6\frac{5}{9}+x-\frac{117}{8}\right):\left(12\frac{1}{9}\right)=0\)
\(\left(\frac{6.9+5}{9}+x-\frac{117}{8}\right):\frac{12.9+1}{9}=\frac{1}{2}\)
( . là nhân nha)
\(\left(\frac{59}{9}-\frac{117}{8}+x\right):\frac{109}{9}=\frac{1}{2}\)
\(\frac{59}{9}-\frac{117}{8}+x=\frac{1}{2}\cdot\frac{109}{9}\)
\(\frac{59}{9}-\frac{117}{8}+x=\frac{109}{18}\)
\(x=\frac{109}{18}-\frac{59}{9}+\frac{117}{8}\)
\(x=\frac{113}{8}\)
( \(\left(y+\frac{1}{3}\right)+\left(y+\frac{2}{9}\right)+\left(y+\frac{1}{27}\right)+\left(y+\frac{1}{81}\right)=\frac{56}{81}\)
\(y+\frac{1}{3}+y+\frac{2}{9}+y+\frac{1}{27}+y+\frac{1}{81}=\frac{56}{81}\)
\(4y+\frac{1}{3}+\frac{2}{9}+\frac{1}{27}+\frac{1}{81}=\frac{56}{81}\)
\(4y+\frac{49}{81}=\frac{56}{81}\)
\(4y=\frac{7}{81}\)
y = 7/81:4
y = 7/324
120 : [ 15 - (3 . y - 2 ) ] = 24
15 - (3 . y - 2) = 120 : 24
15 - (3 . y - 2) = 5
3 . y - 2 = 15 - 5
3 . y - 2 = 10
3 . y = 10 + 2
3 . y = 12
y = 12 : 3
y = 4
Vậy y = 4
120: [15 - (3.y - 2)] = 24
<=> 15 -(3y- 2) = 120: 24
<=> 15 - (3y-2) = 5
<=> 3y-2 = 10
<=> 3y = 12
<=> y = 4
Vậy y= 4
Chúc bn học tốt!