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Câu 1:
\(A=\frac{\left(1+2+3+...+100\right)x\left(101x102-101x101-51-50\right)}{2+4+6+8+...+2048}\)
\(A=\frac{\left(1+2+3+...+100\right)x\left(101x\left(102-101\right)-\left(50+51\right)\right)}{2+4+6+8+...+2048}\)
\(A=\frac{\left(1+2+3+...+100\right)x\left(101-101\right)}{2+4+6+8+...+2048}\)
\(A=\frac{\left(1+2+3+...+100\right)x0}{2+4+6+8+...+2048}\)
\(A=0\)
Ta có:Số số hạng từ 2 đến 101 là:
(101-2):1+1=100(số hạng)
Do đó từ 2 đến 101 có số cặp là:
100:2=50(cặp)
\(B=\frac{101+100+99+...+3+2+1}{101-100+99-98+3-2+1}\)
\(B=\frac{5151}{51}\)
\(B=101\)
Câu 2:
a)697:\(\frac{15x+364}{x}\)=17
\(\frac{15x+364}{x}\)=697:17
\(\frac{15x+364}{x}\)=41
15x+364=41x
41x-15x=364
26x=364
x=14
Vậy x=14
b)92.4-27=\(\frac{x+350}{x}+315\)
\(\frac{x+350}{x}+315\)=341
\(\frac{x+350}{x}\)=26
x+350=26
x=26-350
x=-324
Vậy x=-324
c, 720 : [ 41 - ( 2x -5)] = 40
[ 41 - ( 2x -5)] =720:40
[ 41 - ( 2x -5)] =18
2x-5=41-18
2x-5=23
2x=28
x=14
Vậy x=14
d, Số số hạng từ 1 đến 100 là:
(100-1):1+1=100(số hạng)
Tổng dãy số là:
(100+1)x100:2=5050
Mà cứ 1 số hạng lại có 1x suy ra có 100x
Ta có:(x+1) + (x+2) +...+ (x+100) = 5750
(x+x+...+x)+(1+2+...+100)=5750
100x+5050=5750
100x=700
x=7
Vậy x=7
\(\left(2\frac{4}{5}x-50\right)\div\frac{2}{3}=51\)
\(\frac{14}{5}x-50=51\times\frac{2}{3}\)
\(\frac{14}{5}x-50=34\)
\(\frac{14}{5}x=84\)
\(x=84\times\frac{5}{14}\)
\(x=30\)
\(a,\left(2\frac{4}{5}.x-50\right):\frac{2}{3}=51\)
\(\left(\frac{14}{5}.x-50\right):\frac{2}{3}=51\)
\(\frac{14}{5}.x-50=51\times\frac{2}{3}\)
\(\frac{14}{5}.x-50=34\)
\(\frac{14}{5}.x=34+50\)
\(\frac{14}{5}.x=84\)
\(x=84:\frac{14}{5}\)
\(x=\frac{405}{14}\)
\(a,\frac{x+1}{65}+\frac{x+2}{64}=\frac{x+3}{63}+\frac{x+4}{62}\)
\(\Rightarrow\left[\frac{x+1}{65}+1\right]+\left[\frac{x+2}{64}+1\right]=\left[\frac{x+3}{63}+1\right]+\left[\frac{x+4}{62}+1\right]\)
\(\Rightarrow\frac{x+1+65}{65}+\frac{x+2+64}{64}=\frac{x+3+63}{63}+\frac{x+4+62}{62}\)
\(\Rightarrow\frac{x+66}{65}+\frac{x+66}{64}=\frac{x+66}{63}+\frac{x+66}{62}\)
\(\Rightarrow\frac{x+66}{65}+\frac{x+66}{64}=\frac{x+66}{63}+\frac{x+66}{62}=0\)
\(\Rightarrow\left[x+66\right]\left[\frac{1}{65}+\frac{1}{64}-\frac{1}{63}+\frac{1}{62}\right]=0\)
Mà \(\frac{1}{65}+\frac{1}{64}-\frac{1}{63}+\frac{1}{62}\ne0\)
\(\Rightarrow x+66=0\)
\(\Rightarrow x=0-66=-66\)
Auto làm nốt câu b
a, Cộng cả 2 vế với 2
Ta có \(\frac{x+1}{64}+\frac{x+2}{63}+2=\frac{x+3}{62}+\frac{x+4}{61}+2\)
\(\left(\frac{x+1}{64}+\frac{64}{64}\right)+\left(\frac{x+2}{63}+\frac{63}{63}\right)=\left(\frac{x+3}{62}+\frac{62}{62}\right)+\left(\frac{x+4}{61}+\frac{61}{61}\right)\)
=> \(\frac{x+65}{64}+\frac{x+65}{63}=\frac{x+65}{62}+\frac{x+65}{61}\)\(\)
=> \(\frac{x+65}{64}+\frac{x+65}{63}-\frac{x+65}{62}-\frac{x+65}{61}=0\)
=> \(\left(x+65\right)\left(\frac{1}{64}+\frac{1}{63}-\frac{1}{62}-\frac{1}{61}\right)=0\)
Do \(\frac{1}{64}+\frac{1}{63}-\frac{1}{62}-\frac{1}{61}\ne0\)=> \(x+65=0\)
=> \(x=-65\)
b , Lm tương tự như Câu a
Chúc bn hok tốt
a, \(2.x^x=10.3^{12}+8.27^4\)
\(2.x^x=10.3^{12}+8.3^{12}\)
\(2.x^x=3^{12}.\left(10+8\right)\)
\(2.x^x=3^{12}.18\)
\(2.x^x=3^{12}.2.3^3\)
\(2.x^x=3^{15}.2\)
\(x^x=3^{15}\)( Hình như sai đề )
b,\(3^{2x+2}=9^{x+3}\)
\(3^{2x+2}=3^{2x+3}\)
a) \(\frac{4}{3}x-1=\frac{x}{5}\)
=> \(\frac{4}{3}x-\frac{1}{5}x=1\)
=> \(\frac{17}{15}x=1\)
=> \(x=1:\frac{17}{15}=\frac{15}{17}\)
b) \(x+50\%=\frac{4\left(x+1\right)}{3}-\frac{1}{3}\)
=> \(x+\frac{1}{2}=\frac{4x+4-1}{3}\)
=> \(\frac{2x+1}{2}=\frac{4x+3}{3}\)
=> \(\left(2x+1\right).3=2.\left(4x+3\right)\)
=> \(6x+3=8x+6\)
=> \(6x-8x=6-3\)
=> \(-2x=3\)
=> \(x=3:\left(-2\right)=-\frac{3}{2}\)
\(a,\frac{4}{3}x-1=\frac{x}{5}\)
\(\Rightarrow\frac{4}{3}x=\frac{x}{5}+1\)
\(\Rightarrow\frac{4x}{3}=\frac{x+5}{5}\)
\(\Rightarrow20x=3x+15\)
\(\Rightarrow17x=15\)
\(\Rightarrow x=\frac{15}{17}\)
\(b,x+50\%=\frac{4\left(x+1\right)}{3}-\frac{1}{3}\)
\(\Rightarrow x+\frac{1}{2}=\frac{4x+4-1}{3}\)
\(\Rightarrow\frac{2x+1}{2}=\frac{4x+3}{3}\)
\(\Rightarrow3\left(2x+1\right)=\left(4x+3\right).2\)
\(\Rightarrow6x+3=8x+6\)
\(\Rightarrow2x=-3\)
\(\Rightarrow x=-\frac{3}{2}\)
\(c,-200\%.x+\frac{4}{3}=\frac{7}{4}\left(x+1\right)\)
\(\Rightarrow-2x+\frac{4}{3}=\frac{7}{4}x+\frac{7}{4}\)
\(\Rightarrow\frac{-6x+4}{3}=\frac{7x+7}{4}\)
\(\Rightarrow4\left(-6x+4\right)=\left(7x+7\right)3\)
\(\Rightarrow-24x+16=21x+21\)
\(\Rightarrow45x=-5\)
\(\Rightarrow x=-\frac{1}{7}\)
a)\(\frac{-2}{3}.x+\frac{1}{5}=\frac{3}{10}\)
\(\frac{-2}{3}x=\frac{3}{10}-\frac{1}{5}=\frac{1}{10}\)
\(\frac{-2}{3}x=\frac{1}{10}\)
\(x=\frac{1}{10}\div\frac{-2}{3}=\frac{-3}{20}\)
b)\(\left(50\%.x+2\frac{1}{4}\right).\frac{-2}{3}=\frac{17}{6}\)
\(50\%.x+2\frac{1}{4}=\frac{17}{6}\div\frac{-2}{3}\)\(=\frac{-17}{4}\)
\(50\%.x+2\frac{1}{4}=\frac{-17}{4}\)
\(50\%.x=\frac{-17}{4}-2\frac{1}{4}=\frac{-17}{4}-\frac{9}{4}=\frac{-26}{4}=\frac{-13}{2}\)
\(50\%.x=\frac{-13}{2}\)
\(x=\frac{-13}{2}\div50\%=-13\)
\(\frac{-2}{3}\)\(\hept{\begin{cases}.\\.\\.\end{cases}ads}\)
x2+x3+x4+...+x100=50
Có số số hạng lũy thừa x là: 100-2+1=99(số)
Có số nhóm 3 ghép thành là:99:3=33
Ta có:x2+x3+x4+...+x100=50
<=>(x2+x3+x4)+...+(x98+x99+x100)=50
ôi suy nghĩ cái