tính A=(2+1/315) x (1/651) - (1/105) x (3+650/651) - 4/(315x651) - (4/105)
1) tìm giá trị của biểu thức A ;
A=2/1/315 x 1/650 - 1/105 x3/650/651 - 4/315x651 + 4/105
2) Tính giá trị của biểu thức sau ;
B=x^5 -15.x^4 +16.x^3 -29.x^2 +13.x . Tại x=14
\(2\frac{1}{315}.\frac{1}{651}-\frac{1}{105}.3\frac{650}{651}-\frac{4}{315.651}+\frac{4}{105}\)
\(=\left(2+\frac{1}{315}\right).\frac{1}{651}-\frac{1}{105}.\left(3+\frac{650}{651}\right)-\frac{4}{315.651}+\frac{4}{105}\)
\(=2.\frac{1}{651}+\frac{1}{315}.\frac{1}{651}-\frac{1}{105}.3+\frac{1}{105}.\frac{650}{651}-\frac{4}{315.651}+\frac{4}{105}\)
\(=\frac{2}{651}+\frac{1}{315.651}-\frac{3}{105}-\frac{650}{105.651}-\frac{4}{315.651}+\frac{4}{105}\)
\(=\frac{2}{615}+\left(\frac{1}{315.651}-\frac{4}{315.651}\right)+\left(\frac{-3}{105}+\frac{4}{105}\right)-\frac{650}{105.651}\)
\(=\frac{2}{651}-\frac{3}{315.651}+\frac{1}{105}-\frac{650}{105.651}\)
\(=\left(\frac{2}{651}+\frac{1}{105}\right)-\frac{650}{105.651}-\frac{3}{315.651}\)
\(=\left(\frac{2.105}{105.651}+\frac{651}{105.651}\right)-\frac{650}{105.651}-\frac{3}{315.651}\)
\(=\frac{211}{105.651}-\frac{3}{315.651}\)
\(=\frac{1}{651}.\left(\frac{211}{105}-\frac{3}{315}\right)\)
\(=\frac{1}{651}.\left(\frac{633}{315}-\frac{3}{315}\right)\)
\(=\frac{1}{651}.2\)
\(=\frac{2}{651}\)
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
2.(1/315).(1/651)-(1/105).3.(650/651)-(4/315.651)+(4/105)
Tính giá trị biểu thức
=\(\frac{2.1.1}{315.651}-\frac{1.3.650}{105.651}-\frac{4.651}{315}+\frac{4}{105}\)
=\(\frac{2}{315.651}-\frac{650.3}{105.651}-\frac{4.651}{315}-\frac{4}{105}\)
=
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 :
\(A=2\dfrac{1}{315}.\dfrac{1}{651}-\dfrac{1}{105}.3\dfrac{650}{651}-\dfrac{4}{315-651}+\dfrac{4}{105}\)
\(=\left(2+\dfrac{1}{3.105}\right).\dfrac{1}{651}-\dfrac{1}{105}\left(3+\dfrac{651-1}{651}\right)+\dfrac{4}{105}\)
\(=\left(2+\dfrac{1}{3}.\dfrac{1}{105}\right).\dfrac{1}{651}-\dfrac{1}{105}\left(4-\dfrac{1}{651}\right)-\left(\dfrac{3}{3.105.651}+\dfrac{1}{3.105.651}\right)+\dfrac{4}{105}\)
\(=\left(2+\dfrac{1}{3}x\right)y-x\left(4-y\right)-xy-\dfrac{1}{3}xy+4x\)
\(=2y+\dfrac{1}{3}xy-4x+xy-xy-\dfrac{1}{3}xy+4x\)
\(=2y\)
\(=\dfrac{2}{651}\)
Có sửa lại đề 1 chút á :>
Tính \(\:A=2\frac{1}{315}\cdot\frac{1}{651}-\frac{1}{105}\cdot3\frac{650}{651}-\frac{4}{315\cdot651}+\frac{4}{105}\)
dat \(\frac{1}{315}=a,\frac{1}{651}=b\) \(\frac{1}{105}=c\)
A= 2ab-c(3+1-b)-4ab+4c=2ab-4c+bc-4ab+4c=bc-2ab tự giải tiếp nhé
Tính\(2\frac{1}{315}\cdot\frac{1}{651}-\frac{1}{105}\cdot3\frac{650}{651}-\frac{4}{315\cdot651}+\frac{4}{105}\)
Gợi ý:
Đặt: \(\frac{1}{105}=a;\) \(\frac{1}{651}=b\)
Như vậy \(\frac{1}{315}=3a;\)\(\frac{650}{651}=1-b\)
Sau đó thay vào biểu thức ban đầu, bạn tự làm tiếp nhé
Tính \(A=2\frac{1}{315}.\frac{1}{651}-\frac{1}{105}.3\frac{650}{651}-\frac{4}{315.651}+\frac{4}{105}\)
Đặt \(a=\frac{1}{315}\), \(b=\frac{1}{651}\)ta có :
\(A=\left(2+a\right)\cdot b-3a\left(3+1-b\right)-4ab+12a\)
\(\Rightarrow A=2b+ab-12a+3ab-4ab+12a\)
\(\Rightarrow A=2b=\frac{2}{651}\)
1) Cho
\(B=2\frac{1}{315}.\frac{1}{651}-\frac{1}{105}.3\frac{650}{651}-\frac{4}{315.651}+\frac{1}{105}\)
Đặt
\(x=\frac{1}{315};y=\frac{1}{651};z=\frac{1}{105}\)
Hãy tính giá trị của B