651 + 845
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2frac{1}{315}.frac{1}{651}-frac{1}{105}.3frac{650}{651}-frac{4}{315.651}+frac{4}{105}left(2+frac{1}{315}right).frac{1}{651}-frac{1}{105}.left(3+frac{650}{651}right)-frac{4}{315.651}+frac{4}{105}2.frac{1}{651}+frac{1}{315}.frac{1}{651}-frac{1}{105}.3+frac{1}{105}.frac{650}{651}-frac{4}{315.651}+frac{4}{105}frac{2}{651}+frac{1}{315.651}-frac{3}{105}-frac{650}{105.651}-frac{4}{315.651}+frac{4}{105}frac{2}{615}+left(frac{1}{315.651}-frac{4}{315.651}right)+left(frac{-3}{105}+frac{4}{105}right)-frac{6...
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\(2\frac{1}{315}.\frac{1}{651}-\frac{1}{105}.3\frac{650}{651}-\frac{4}{315.651}+\frac{4}{105}\)
\(=\left(2+\frac{1}{315}\right).\frac{1}{651}-\frac{1}{105}.\left(3+\frac{650}{651}\right)-\frac{4}{315.651}+\frac{4}{105}\)
\(=2.\frac{1}{651}+\frac{1}{315}.\frac{1}{651}-\frac{1}{105}.3+\frac{1}{105}.\frac{650}{651}-\frac{4}{315.651}+\frac{4}{105}\)
\(=\frac{2}{651}+\frac{1}{315.651}-\frac{3}{105}-\frac{650}{105.651}-\frac{4}{315.651}+\frac{4}{105}\)
\(=\frac{2}{615}+\left(\frac{1}{315.651}-\frac{4}{315.651}\right)+\left(\frac{-3}{105}+\frac{4}{105}\right)-\frac{650}{105.651}\)
\(=\frac{2}{651}-\frac{3}{315.651}+\frac{1}{105}-\frac{650}{105.651}\)
\(=\left(\frac{2}{651}+\frac{1}{105}\right)-\frac{650}{105.651}-\frac{3}{315.651}\)
\(=\left(\frac{2.105}{105.651}+\frac{651}{105.651}\right)-\frac{650}{105.651}-\frac{3}{315.651}\)
\(=\frac{211}{105.651}-\frac{3}{315.651}\)
\(=\frac{1}{651}.\left(\frac{211}{105}-\frac{3}{315}\right)\)
\(=\frac{1}{651}.\left(\frac{633}{315}-\frac{3}{315}\right)\)
\(=\frac{1}{651}.2\)
\(=\frac{2}{651}\)
A= 631/315 .1/651 -1/105 .2603/651 -4/315.651 +4/105
A=\(\dfrac{631}{315}.\dfrac{1}{651}-\dfrac{1}{105}.2\dfrac{603}{651}-\dfrac{4}{315.651}+\dfrac{4}{105}\)
Đặt \(a=\dfrac{1}{105}\);b=\(\dfrac{1}{651}\)
Rồi giải ra
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tính A=(2+1/315) x (1/651) - (1/105) x (3+650/651) - 4/(315x651) - (4/105)
2+1÷315.1÷615-1÷105.3+650÷651-4÷315×651+4÷105
845 : 7 = ?
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2.(1/315).(1/651)-(1/105).3.(650/651)-(4/315.651)+(4/105)
Tính giá trị biểu thức
=\(\frac{2.1.1}{315.651}-\frac{1.3.650}{105.651}-\frac{4.651}{315}+\frac{4}{105}\)
=\(\frac{2}{315.651}-\frac{650.3}{105.651}-\frac{4.651}{315}-\frac{4}{105}\)
=
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tính giá trị của biểu thức :
M=( 2+1/315 ) * 1/651 -1/105 * ( 3+ 650/651 ) -4 / 315*615 +4/105
Đặt \(\frac{1}{315}=a\)và\(\frac{1}{651}=b\) rồi biến đổi M
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cho mk hỏi làm sao từ \(A=2.\frac{1}{315}.\frac{1}{651}-\frac{1}{105}.3\frac{650}{651}-\frac{4}{315.651}+\frac{4}{105}\)
=> \(A=\left(2+\frac{1}{315}\right).\frac{1}{651}.......\)
Bạn chú ý rằng 315= 3.105 và 650=651-1 nha
Thế 2 cái đó vào A là ra thoi !!
Cho mình hỏi cái chỗ \(3\frac{650}{651}\)là hỗn số hay phép nhân thế
Roi mình giải lun cho
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1) Tìm x
4x(x-1)-x(4x-2) = 5(x-3)
2. Tính hợp lý
A = \(2\dfrac{1}{315}-\dfrac{1}{651}-\dfrac{1}{105}.3\dfrac{650}{651}-\dfrac{4}{315-651} +\dfrac{4}{105}\)
1/ \(4x\left(x-1\right)-x\left(4x-2\right)=5\left(x-3\right)\)
\(\Leftrightarrow4x^2-4x-4x^2+2x=5x-15\)
\(\Leftrightarrow-2x=5x-15\)
\(\Leftrightarrow7x=-15\)
\(\Leftrightarrow x=-\dfrac{15}{7}\)
Vậy ..
2/ Đặt : \(\dfrac{1}{105}=x;\dfrac{1}{651}=y\left(x,y>0\right)\)
Ta có :
\(A=2\dfrac{1}{315}.\dfrac{1}{651}-\dfrac{1}{105}.3\dfrac{650}{651}-\dfrac{4}{315-651}+\dfrac{4}{105}\)
\(=\left(2+\dfrac{1}{3.105}\right).\dfrac{1}{651}-\dfrac{1}{105}\left(3+\dfrac{651-1}{651}\right)+\dfrac{4}{105}\)
\(=\left(2+\dfrac{1}{3}.\dfrac{1}{105}\right).\dfrac{1}{651}-\dfrac{1}{105}\left(4-\dfrac{1}{651}\right)-\left(\dfrac{3}{3.105.651}+\dfrac{1}{3.105.651}\right)+\dfrac{4}{105}\)
\(=\left(2+\dfrac{1}{3}x\right)y-x\left(4-y\right)-xy-\dfrac{1}{3}xy+4x\)
\(=2y+\dfrac{1}{3}xy-4x+xy-xy-\dfrac{1}{3}xy+4x\)
\(=2y\)
\(=\dfrac{2}{651}\)
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