(1-1/2) x (1-1/3) x ......................x (1-1/10)
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a: =>14x+20+5=6x-9-9x
=>14x+25=-3x-9
=>17x=-34
=>x=-2
b: =>\(2x^2-30x+2x-30=2x^2+10x-10x-50\)
=>-28x-30=-50
=>-28x=-20
=>x=20/28=5/7
c: =>2x+x^3-x=x^3+1
=>x=1
d: =>x^3-3x^2+3x-1-x(x^2+2x+1)=10x-2x^2-11x-22
=>x^3-3x^2+3x-1-x^3-2x^2-x=-2x^2-x-22
=>-5x^2+2x-1+2x^2+x+22=0
=>-3x^2+3x+21=0
=>x^2-x-7=0
=>\(x=\dfrac{1\pm\sqrt{29}}{2}\)
Câu 1 :
a, \(\frac{3\left(2x+1\right)}{4}-\frac{5x+3}{6}=\frac{2x-1}{3}-\frac{3-x}{4}\)
\(\Leftrightarrow\frac{6x+3}{4}+\frac{3-x}{4}=\frac{2x-1}{3}+\frac{5x+3}{6}\)
\(\Leftrightarrow\frac{5x+6}{4}=\frac{9x+1}{6}\Leftrightarrow\frac{30x+36}{24}=\frac{36x+4}{24}\)
Khử mẫu : \(30x+36=36x+4\Leftrightarrow-6x=-32\Leftrightarrow x=\frac{32}{6}=\frac{16}{3}\)
tương tự
\(\frac{19}{4}-\frac{2\left(3x-5\right)}{5}=\frac{3-2x}{10}-\frac{3x-1}{4}\)
\(< =>\frac{19.5}{20}-\frac{8\left(3x-5\right)}{20}=\frac{2\left(3-2x\right)}{20}-\frac{5\left(3x-1\right)}{20}\)
\(< =>95-24x+40=6-4x-15x+5\)
\(< =>-24x+135=-19x+11\)
\(< =>5x=135-11=124\)
\(< =>x=\frac{124}{5}\)
Câu 1 : \(\frac{10}{3}\cdot x-\frac{1}{2}=\frac{1}{10}\)
\(\Rightarrow\frac{10}{3}\cdot x=\frac{3}{5}\Rightarrow x=\frac{9}{50}\)
Câu 2 : \(\frac{1}{3}\cdot x+\frac{1}{3}\cdot x=\frac{1}{8}\Rightarrow x\cdot\left(\frac{1}{3}+\frac{1}{3}\right)=\frac{1}{8}\)
\(x\cdot\frac{2}{3}=\frac{1}{8}\Rightarrow x=\frac{3}{16}\)
Câu 3 : \(\left[3x-1\right]\left[\frac{1}{2}x-\frac{2}{5}\right]=0\)
\(\Rightarrow3x-1=0\)hoặc \(\frac{1}{2}x-\frac{2}{5}=0\)
( vô lí ) x = 4/5
a: Ta có: \(\left(x+1\right)^3-\left(x+2\right)\left(x-1\right)^2-3\left(x-3\right)\left(x+3\right)=5\)
\(\Leftrightarrow x^3+3x^2+3x+1-\left(x+2\right)\left(x^2-2x+1\right)-3\left(x^2-9\right)=5\)
\(\Leftrightarrow x^3+3x^2+3x+1-\left(x^3-2x^2+x+2x^2-4x+2\right)-3\left(x^2-9\right)=5\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x-2-3x^2+9=5\)
\(\Leftrightarrow6x=-3\)
hay \(x=-\dfrac{1}{2}\)
b: Ta có: \(\left(x+1\right)^3+\left(x-1\right)^3=\left(x+2\right)^3+\left(x-2\right)^3\)
\(\Leftrightarrow x^3+3x^2+3x+1+x^3-3x^2+3x-1=x^3+6x^2+12x+8+x^3-6x^2+12x-8\)
\(\Leftrightarrow2x^3+6x=2x^3+24x\)
\(\Leftrightarrow x=0\)
c: Ta có: \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-10\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-1=-10\)
\(\Leftrightarrow12x=-11\)
hay \(x=-\dfrac{11}{12}\)
a) \(\dfrac{1}{2}+x=\dfrac{5}{6}\)
\(\Rightarrow x=\dfrac{5}{6}-\dfrac{1}{2}=\dfrac{5}{6}-\dfrac{3}{6}=\dfrac{2}{6}=\dfrac{1}{3}\)
b) \(x+\dfrac{1}{4}=\dfrac{3}{4}\)
\(\Rightarrow x=\dfrac{3}{4}-\dfrac{1}{4}=\dfrac{2}{4}=\dfrac{1}{2}\)
c) \(x-\dfrac{1}{5}=\dfrac{3}{10}\)
\(\Rightarrow x=\dfrac{3}{10}+\dfrac{1}{5}=\dfrac{3}{10}+\dfrac{2}{10}=\dfrac{5}{10}=\dfrac{1}{2}\)
d) \(\dfrac{5}{6}-x=\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{5}{6}-\dfrac{1}{3}=\dfrac{5}{6}-\dfrac{2}{6}=\dfrac{3}{6}=\dfrac{1}{2}\)
e) \(\dfrac{3}{10}+x=\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{1}{2}-\dfrac{3}{10}=\dfrac{5}{10}-\dfrac{3}{10}=\dfrac{2}{10}=\dfrac{1}{5}\)
g) \(x+\dfrac{1}{4}=\dfrac{3}{8}\)
\(\Rightarrow x=\dfrac{3}{8}-\dfrac{1}{4}=\dfrac{3}{8}-\dfrac{2}{8}=\dfrac{1}{8}\)
a) 12+x=5612+x=56
⇒x=56−12=56−36=26=13⇒x=56−12=56−36=26=13
b) x+14=34x+14=34
⇒x=34−14=24=12⇒x=34−14=24=12
c) x−15=310x−15=310
⇒x=310+15=310+210=510=12⇒x=310+15=310+210=510=12
d) 56−x=1356−x=13
⇒x=56−13=56−26=36=12⇒x=56−13=56−26=36=12
e) 310+x=12310+x=12
⇒x=12−310=510−310=210=15⇒x=12−310=510−310=210=15
g) x+14=38x+14=38
⇒x=38−14=38−28=18⇒x=38−14=38−28=18
Đọc tiếp
\(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot...\cdot\left(1-\frac{1}{10}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot...\cdot\frac{9}{10}\)
\(=\frac{1\cdot2\cdot...\cdot9}{2\cdot3\cdot..\cdot10}\)
\(=\frac{1}{10}\)
thanks