Tính giá trị của : \(\dfrac{1}{3\text{x}4}\)+\(\dfrac{1}{4\text{x}5}\)+\(\dfrac{1}{5\text{x}6}\)+...+\(\dfrac{1}{38\text{x}39}\)
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\(\dfrac{12^3.18^2}{24^2}=\dfrac{12^3.6^2.3^2}{6^2.4^2}=\dfrac{4^3.3^3.3^2}{4^2}=4.3^3.3^2=4.3^5=972\)
\(\dfrac{12^3.18^2}{24^2}=\dfrac{1728.324}{576}=\dfrac{559872}{576}=972\)
Vậy giá trị cần tìm là 972
\(\left[\left(6\times x-72\right):2-84\right]\times28=5628\)
\(\left(6\times x-72\right):2-84=5628:28\)
\(\left(6.x-72\right):2-84=201\)
\(\left(6.x-72\right):2=201+84\)
\(\left(6.x-72\right):2=285\)
\(6.x-72=285.2\)
\(6.x-72=570\)
\(6.x=570+72\)
\(6.x=642\)
\(x=642:6\)
\(x=107\)
[(6xX-72):2-84]x28=5628
⇒ (6xX-72):2-84=5629:28=201
⇒ (6xX-72):2=201+84=285
⇒ 6xX-72=285x2=570
⇒ 6xX=570+72=642
⇒ X= 642:6=107
⇒ X=107
\(\dfrac{2\cdot3+4\cdot6+14\cdot21}{3\cdot5+6\cdot10+21\cdot35}=\dfrac{2\left(3+2\cdot3+7\cdot21\right)}{3\left(5+2\cdot10+7\cdot35\right)}=\dfrac{2\cdot3\left(1+2\cdot3+7\cdot5\right)}{3\cdot5\left(1+2\cdot3+7\cdot5\right)}=\dfrac{2}{5}\)
\(\dfrac{2.3+4.6+14.21}{3.5+6.10+21.35}\)
\(=\dfrac{6+24+294}{15+60+735}\)
\(=\dfrac{324}{810}\)
\(=\dfrac{2}{5}\)
\(\left(2600+6400\right)-3.X=1200\)
\(9000-3.X=1200\)
\(3.X=9000-1200\)
\(3.X=7800\)
\(X=7800:3\)
\(X=2600\)
(2600+6400)-3xX=1200
⇒ 9000-3xX=1200
⇒ 3xX=9000-1200=7800
⇒ X=7800:3=2600
⇒ X=2600
\(99^{3+1+3}.\left(99^2+99\right)=99^7.\left(99^2+99\right)=99^7.99^2+99^7.99=99^{99}+99^{98}\)
A = \(x\) - 2\(x\) + 3
A = - \(x\) + 3
|\(x\)| = 0,5
\(\left[{}\begin{matrix}x=0,5\\x=-0,5\end{matrix}\right.\)
Thay \(x=0,5\) vào A ta có: A = -0,5 + 3 = 2,5
Thay \(x\) = -,05 vào A ta có: A = -(-0,5) + 3 = 3,5
219-7x(x-1)=100
⇒ 7x(x-1)=219-100
⇒ 7x(x-1)=119
⇒ x-1=119:7
⇒ x-1=17
⇒ x=17+1
⇒ x=18
219-7(x+1)=100
7(x+1)=219-100
7(x+1)=119
x+1=119:7
x+1=17
x=17-1
x=16
(x:2-1)x13-12=1
⇒ (x:2-1)x13=1+12
⇒ (x:2-1)x13=13
⇒ x:2-1=13:13
⇒ x:2-1=1
⇒ x:2=1+1
⇒ x:2=2
⇒ x=2x2
⇒ x=4
\(\left(X:2-1\right).13-12=1\)
\(\left(X:2-1\right).13=1+12\)
\(\left(X:2-1\right).13=13\)
\(X:2-1=13:13\)
\(X:2-1=1\)
\(X:2=1+1\)
\(X:2=2\)
\(X=2.2\)
\(X=4\)
\(\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+...+\dfrac{1}{38\times39}\)
\(=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...\dfrac{1}{38}-\dfrac{1}{39}\)
\(=\dfrac{1}{3}-\dfrac{1}{39}\)
\(=\dfrac{13}{39}-\dfrac{1}{39}\)
\(=\dfrac{12}{39}=\dfrac{4}{13}\)
`1/(3xx4)+1/(4xx5)+1/(5xx6)+...+1/(38xx39)`
`=1/3-1/4+1/4-1/5+1/5-1/6+...+1/38-1/39`
`=1/3-1/39`
`=4/13`