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\(\dfrac{1}{6}x+\dfrac{1}{12}x+\dfrac{1}{20}x+...+\dfrac{1}{2450}x=1\)
\(x\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{2450}\right)\)=1
\(x\left(\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+...+\dfrac{1}{49\times50}\right)\)=1
\(x\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=1\)
\(x\left(\dfrac{1}{2}-\dfrac{1}{50}\right)=1\)
\(x\times\)\(\dfrac{12}{25}=1\)
\(\Rightarrow x=1\div\dfrac{12}{25}\)
\(x=1\times\dfrac{25}{12}=\dfrac{25}{12}\)
vậy \(x=\dfrac{25}{12}\)
vậy \(x=2\)\(x=2\)\(\Rightarrow\left(\dfrac{20}{9}-x\right)=\dfrac{1}{3}-\dfrac{1}{9}\)\(\left(2\dfrac{2}{9}-x\right)\)=\(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{72}\)
\(\Rightarrow\left(\dfrac{20}{9}-x\right)=\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{72}\)\(\Rightarrow\left(\dfrac{20}{9}-x\right)=\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+...+\dfrac{1}{8\times9}\)\(\Rightarrow\left(\dfrac{20}{9}-x\right)=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{8}-\dfrac{1}{9}\)
a: \(\Leftrightarrow3x+7\in\left\{1;-1;3;-3;11;-11;33;-33\right\}\)
hay \(x=-6\)
b: \(\Leftrightarrow3x-1\in\left\{1;-1;7;-7\right\}\)
hay \(x\in\left\{0;-2\right\}\)
cách 1: phân tích ra ước
cách 2 áp dụng 7 hằng đẳng thức nhân tung ra
2: 12-10x=25-30x
=>20x=13
=>x=13/20
3: \(3\left(2x+3\right)-2\left(4x-5\right)=10x+21\)
=>6x+9-8x+10=10x+21
=>10x+21=-2x+19
=>12x=-2
=>x=-1/6
4: \(\Leftrightarrow25x-15-6x+12=11-5x\)
=>19x-3=11-5x
=>24x=14
=>x=7/12
5: \(\Leftrightarrow8-12x-5+10x=4-6x\)
=>4-6x=-2x+3
=>-4x=-1
=>x=1/4
6: \(\Leftrightarrow32x-24-6+9x=13-40x\)
=>41x-30=13-40x
=>81x=43
=>x=43/81
7: \(\Leftrightarrow10x-5+20x=5x-11\)
=>30x-5=5x-11
=>25x=-6
=>x=-6/25
Gọi biểu thức đó là A
\(A=\dfrac{1}{x}.\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+..+\dfrac{1}{49.50}\right)=1\)
\(\Rightarrow A=\dfrac{1}{x}.\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{49.50}\right)=1\)
Ta có công thức : \(\dfrac{a}{b.c}=\dfrac{a}{c-b}.\left(\dfrac{1}{b}-\dfrac{1}{c}\right)\)
Dựa vào công thức trên ta có :
\(\dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3}\)
\(\dfrac{1}{3.4}=\dfrac{1}{3}-\dfrac{1}{4}\)
......................
\(\dfrac{1}{49.50}=\dfrac{1}{49}-\dfrac{1}{50}\)
\(\Rightarrow A=\dfrac{1}{x}.\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+.....+\dfrac{1}{49}-\dfrac{1}{50}\right)=1\)
\(\Rightarrow A=\dfrac{1}{x}.\left(\dfrac{1}{2}-\dfrac{1}{50}\right)=1\)
\(\Leftrightarrow A=\dfrac{1}{x}.\left(\dfrac{12}{25}\right)=1\)
\(\Rightarrow\) \(\dfrac{1}{x}=A:\dfrac{12}{25}=1:\dfrac{12}{25}=\dfrac{25}{12}\)
Vậy x = 12.
Mink nghĩ vậy, ai thấy đúng thì ủng hộ mink nha !!!