Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) 5 - x + 12 = 4 + x + 1
17 - x = 5 + x
x - (-x) = 17 - 5
2x = 12
x = 6
b) 4x - 5 + (-15) = 3x - 10
4x - 20 = 3x - 10
3x - 4x = -20 + 10
-x = -10
x = 10
a)
5 - x + 12 = 4 + x + 1
17 - x = 5 + x x - (-x)
= 17 - 5 2x
= 12 x
= 6 b)
4x - 5 + (-15)
= 3x - 10 4x - 20
= 3x - 10 3x - 4x
= -20 + 10 -x
= -10 x = 10
\(\left|2x+1\right|-x=5\)
\(\Rightarrow2x+1-x=5\)
\(\Rightarrow x+1=5\)
\(\Rightarrow x=5-1\)
\(\Rightarrow x=4\)
Tương tự có ( x - 1 ) : 2 + 1 số hạng
( x + 1 ) [ ( x - 1 ) : 2 + 1 ] = 72
( x + 1 ) ( x - 1 ) : 2 + ( x + 1 ) = 72
( x + 1 ) ( x - 1 ) + 2 ( x + 1 ) = 144
( x + 1 ) ( x + 1 ) = 144
144 = 12 x 12
Suy ra x + 1 = 12 suy ra x = 11
Ta có:
\(\frac{x-2}{4}=\frac{-16}{2-x}\Leftrightarrow\left(x-2\right)\left(2-x\right)=4.\left(-16\right)\Leftrightarrow-\left(x-2\right)\left(x-2\right)=-64\Leftrightarrow\left(x-2\right)^2=64\)
=> x - 2 = 8 hoặc x - 2 = - 8
=> x = 10 hoặc x = - 6
\(\frac{x-2}{4}=\frac{-16}{2-x}\Rightarrow\left(x-2\right)\left(2-x\right)=4.\left(-16\right)=-64\)
ta thấy x-2 và 2-x là 2 số đối nhau mà -64=8.(-8)
+)TH1: x-2=8 và 2-x=-8
x-2=8=>x=10 nhưng \(2-10\ne-8\)(loại)
+)TH2: x-2=-8 và 2-x=8
x-2=-8=>x=-6 mà 2-(-6)=2+6=8(chọn)
vậy x=-6
\(A=\dfrac{7}{1.9}+\dfrac{7}{9.17}+\dfrac{7}{17.25}+...+\dfrac{7}{81.89}\)
\(\dfrac{8}{7}A=\dfrac{8}{1.9}+\dfrac{8}{9.17}+\dfrac{8}{17.25}+...+\dfrac{8}{81.89}\)
\(\dfrac{8}{7}A=1-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{25}+...+\dfrac{1}{81}-\dfrac{1}{89}\)
\(\dfrac{8}{7}A=1-\dfrac{1}{89}=\dfrac{88}{89}\Rightarrow A=\dfrac{88}{89}:\dfrac{8}{7}=\dfrac{77}{89}\)
\(B=\dfrac{5^2}{1.4}+\dfrac{3^2}{4.7}+\dfrac{3^2}{7.10}+...+\dfrac{3^2}{37.40}\)
\(B=\dfrac{25}{1.4}+\dfrac{9}{4.7}+\dfrac{9}{7.10}+...+\dfrac{9}{37.40}\)
\(\dfrac{1}{3}B=\dfrac{25}{12}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{37.40}\)
\(\dfrac{1}{3}B=\dfrac{25}{12}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{37}-\dfrac{1}{40}\)
\(\dfrac{1}{3}B=\dfrac{25}{12}+\dfrac{1}{4}-\dfrac{1}{40}=\dfrac{277}{120}\Rightarrow B=\dfrac{277}{120}:\dfrac{1}{3}=\dfrac{277}{40}\)
\(A=\dfrac{7}{1.9}+\dfrac{7}{9.17}+\dfrac{7}{17.25}+...+\dfrac{7}{81.89}\)
\(=7\left(\dfrac{8}{1.9}+\dfrac{8}{9.17}+\dfrac{8}{17.25}+...+\dfrac{8}{81.89}\right)\)
\(=7\left(1-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{25}+\dfrac{1}{25}+...+\dfrac{1}{81}-\dfrac{1}{89}\right)\)
\(=7.\left(1-\dfrac{1}{89}\right)=7.\dfrac{88}{89}=\dfrac{616}{89}\)
xet 1/4x7 +1/7x10+...+1/37x40
dat bieu thuc tren la A ta co:
A=1/4x7+1/7x10+...+1/37x40
3A=3/4X7+3/7X10+...+3/37X40
3A=1/4-1/71/7-1/10+...+1/37-1/40
3A=1/4-1/40
3A=10/40-1/40=9/40
A=9/40:3=9/120=3/40
=> (1/4x7+1/7x10+...+1/37x40)-x=4/5
3/40-x=4/5
x=3/40-4/5=-29/40