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Bạn chú ý rằng 315= 3.105 và 650=651-1 nha
Thế 2 cái đó vào A là ra thoi !!
1.
\(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)
\(MC:12\)
Quy đồng :
\(\Rightarrow\frac{3.\left(2x+3\right)}{12}-\left(\frac{2.\left(5x+3\right)}{12}\right)=\frac{3x-4}{12}\)
\(\frac{6x+9}{12}-\left(\frac{10x+6}{12}\right)=\frac{3x-4}{12}\)
\(\Leftrightarrow6x+9-\left(10x+6\right)=3x-4\)
\(\Leftrightarrow6x+9-3x=-4-9+16\)
\(\Leftrightarrow-7x=3\)
\(\Leftrightarrow x=\frac{-3}{7}\)
2.\(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)
\(MC:20\)
Quy đồng :
\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)
\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)
\(\Leftrightarrow30x+15-20=15x-2\)
\(\Leftrightarrow15x=3\)
\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)
\(C=\frac{631}{315}.\frac{1}{651}-\frac{1}{103}.\frac{2603}{651}-\frac{4}{315}.\frac{1}{651}+\frac{4}{105}\)
\(=\frac{1}{651}.\left(\frac{631}{315}-\frac{4}{315}\right)+\frac{2603}{68355}+\frac{4}{105}\)
\(=\frac{1}{651}.\frac{209}{105}+\frac{2603}{68355}+\frac{4}{105}\)
\(=\frac{1}{105}.\left(\frac{209}{651}+\frac{4}{105}\right)+\frac{2603}{68355}\)
\(=105.\frac{167}{465}+\frac{2603}{68355}\)
\(=\frac{1169}{31}+\frac{2603}{68355}=37,74775803\)
Tính nhanh và gọn hết cỡ đc có vậy thôi. Bạn xem lại đề bài nhé 
Đặt \(a=\frac{1}{315}\), \(b=\frac{1}{651}\)ta có :
\(A=\left(2+a\right)\cdot b-3a\left(3+1-b\right)-4ab+12a\)
\(\Rightarrow A=2b+ab-12a+3ab-4ab+12a\)
\(\Rightarrow A=2b=\frac{2}{651}\)
tớ ko bt lm abc , tớ lm d thôi nha , thứ lỗi
\(\frac{5}{2x-3}-\frac{1}{x+2}=\frac{5}{x-6}-\frac{7}{2x-1}\)
\(\frac{3x+13}{2x^2+x-6}=\frac{5}{x-6}+\frac{7}{1-2x}\)
\(\frac{3x+13}{\left(x+2\right)\left(2x-3\right)}=\frac{3x+37}{\left(x-6\right)\left(2x-1\right)}\)
\(\frac{10-9x}{-4x^3+32x^2-51x+18}=0\)
\(\Rightarrow\orbr{\begin{cases}x=-3\\x=\frac{10}{9}\end{cases}}\)
1272 + 146.127 + 732
= 1272 + 2 . 73 .127 + 732
= (127 + 73 ) 2
= 200 2
1) \(63^2-47^2=\left(63+47\right)\left(63-47\right)=110.16=1760\)
2) \(127^2+146.127+73^2=\left(127+73\right)^2=200^2=40000\)
3) \(215^2-105^2=\left(215-105\right)\left(215+105\right)=110.320=35200\)
4) mk chỉnh lại đề:
\(\left(4+1\right)\left(4^2+1\right)\left(4^4+1\right)\left(4^8+1\right)...\left(4^{256}+1\right)\)
\(=\frac{1}{3}\left(4-1\right)\left(4+1\right)\left(4^2+1\right)\left(4^4+1\right)...\left(4^{256}+1\right)\)
\(=\frac{1}{3}\left(4^2-1\right)\left(4^2+1\right)\left(4^4+1\right)...\left(4^{256}+1\right)\)
\(=\frac{1}{3}\left(4^4-1\right)\left(4^4+1\right)...\left(4^{256}+1\right)\)
\(=\frac{1}{3}\left(4^{512}-1\right)\)