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Bài 1:
a: Ta có: \(48751-\left(10425+y\right)=3828:12\)
\(\Leftrightarrow y+10425=48751-319=48432\)
hay y=38007
b: Ta có: \(\left(2367-y\right)-\left(2^{10}-7\right)=15^2-20\)
\(\Leftrightarrow2367-y=1222\)
hay y=1145
Bài 2:
Ta có: \(8\cdot6+288:\left(x-3\right)^2=50\)
\(\Leftrightarrow288:\left(x-3\right)^2=2\)
\(\Leftrightarrow\left(x-3\right)^2=144\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=12\\x-3=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=15\\x=-9\end{matrix}\right.\)
Bài 1:
\(101\cdot125+101\cdot25-101\cdot50\)
\(=101\cdot\left(125+25-50\right)\)
\(=101\cdot100\)
\(=10100\)
Bài 2:
\(76\cdot115+56\cdot24+59\cdot24\)
\(=76\cdot115+24\cdot\left(56+59\right)\)
\(=76\cdot115+24\cdot115\)
\(=115\cdot\left(76+24\right)\)
\(=115\cdot100\)
\(=11500\)
1.
\(\left(\frac{3}{1\times3}+\frac{3}{3\times5}+\frac{3}{5\times7}+...+\frac{3}{97\times99}\right)-x:\frac{3}{2}=\frac{7}{3}\\
\left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{97\times99}\right):\frac{3}{2}-x:\frac{3}{2}=\frac{7}{3}\\\left[\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)-x\right]:\frac{3}{2}=\frac{7}{3}\\
\left(1-\frac{1}{99}\right)-x=\frac{7}{3}\times\frac{3}{2}\\
\frac{98}{99}-x=\frac{7}{2}\\
x=\frac{98}{99}-\frac{7}{2}=\frac{-497}{198}\)
2.\(\frac{x}{y}=\frac{4}{3}\Rightarrow\hept{\begin{cases}x=4a\\y=3a\\x-y=4a-3a=a\end{cases}}\\ \left(x-y\right)^{2015}=5^{2015}\Rightarrow x-y=5\\ \Rightarrow a=5\Rightarrow\hept{\begin{cases}x=4\times5=20\\y=3\times5=15\end{cases}}\)
1) (x - 35) - 120 = 0
x - 35 = 120
x = 120 + 35
x = 155
2) 310 - (118 - x) = 217
118 - x = 310 - 217
118 - x = 93
x = 118 - 93
x = 25
3) 156 - (x + 61) = 82
x + 61 = 156 - 82
x + 61 = 74
x = 74 - 61
x = 13
4) 814 - (x - 305) = 712
x - 305 = 814 - 712
x - 305 = 102
x = 102 + 305 = 407
5) 100 - 7 - (x - 5) = 58
x - 5 = 93 - 58
x - 5 = 35
x = 35 + 5 = 40
6) 12(x - 1) : 3 = 43 + 23
4(x - 1) = 72
x - 1 = 18
x = 18 + 1 = 19
7) 24 + 5x = 75 : 73
24 + 5x = 49
5x = 25
x = 25 : 5 = 5
8) 5(x - 1) : 3 = 43 + 23
\(\dfrac{5}{3}\left(x-1\right)=72\)
x - 1 = \(\dfrac{216}{5}\)
x = 221/5
9) 5(x - 4)2 - 7 = 13
5(x - 4)2 = 20
(x - 4)2 = 4
\(\Rightarrow\left[{}\begin{matrix}x-4=2\\x-4=-2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=6\\x=2\end{matrix}\right.\)
10) (x + 1) + (x + 2) + ... + (x + 30) = 795
=> (x + x + x + ... + x) + (1 + 2 + 3 +...+ 30) = 795 (1)
Đặt A = 1 + 2 + 3 +...+ 30
Số số hạng trong A là: (30 - 1) : 1 + 1 = 30 (số)
Tổng A bằng : (30 + 1).30 : 2 =465
Thay A = 465 vào (1) , ta được:
30x + 465 = 795
=> 30x =330
=> x =11
1: =>x-35=120
=>x=120+35=155
2: =>118-x=310-217=93
=>x=118-93=25
3: =>x+61=156-82=74
=>x=74-61=13
4: =>x-305=814-712=102
=>x=102+305=407
5: =>93-(x-5)=58
=>x-5=35
=>x=40
6: =>4(x-1)=64+8=72
=>x-1=18
=>x=19
7: =>5x+24=49
=>5x=25
=>x=5
8: =>5(x-1):3=4^3+2^3=64+8=72
=>5(x-1)=216
=>x-1=216/5
=>x=221/5
a)
\(\begin{array}{l}\left( {13x{\rm{ }}-{\rm{ }}{{12}^2}} \right):{\rm{ }}5{\rm{ }} = {\rm{ }}5\\13x{\rm{ }}-{\rm{ }}{12^2} = 5.5\\13x{\rm{ }}-{\rm{ }}144 = 25\\13x = 25 + 144\\13x = 169\\x = 13\end{array}\)
Vậy \(x = 13\)
b)
\(\begin{array}{l}3x\left[ {{8^2} - 2.\left( {{2^5} - {\rm{ }}1} \right)} \right]{\rm{ }} = {\rm{ }}2022\\3x\left[ {64 - 2.\left( {32 - {\rm{ }}1} \right)} \right]{\rm{ }} = {\rm{ }}2022\\3x\left[ {64 - 2.31} \right]{\rm{ }} = {\rm{ }}2022\\3x\left( {64 - 62} \right){\rm{ }} = {\rm{ }}2022\\3x.2 = 2022\\6x = 2022\\x = 337\end{array}\)
Vậy \(x = 337.\)
