\(\dfrac{x^2}{2}\)+\(\dfrac{18}{x^2}\)=13*(\(\dfrac{x}{2}\)-\(\dfrac{3}{x}\))
a, \(\dfrac{x-2}{15}+\dfrac{x-3}{14}+\dfrac{x-4}{13}+\dfrac{x-5}{12}=4\)
b, \(\dfrac{x+1}{19}+\dfrac{x+2}{18}+\dfrac{x+3}{17}+...+\dfrac{x+18}{2}+18=0\)
Cảm ơn khi đã giúp mình
a) Ta có: \(\dfrac{x-2}{15}+\dfrac{x-3}{14}+\dfrac{x-4}{13}+\dfrac{x-5}{12}=4\)
\(\Leftrightarrow\dfrac{x-2}{15}-1+\dfrac{x-3}{14}-1+\dfrac{x-4}{13}-1+\dfrac{x-5}{12}-1=0\)
\(\Leftrightarrow\dfrac{x-17}{15}+\dfrac{x-17}{14}+\dfrac{x-17}{13}+\dfrac{x-17}{12}=0\)
\(\Leftrightarrow\left(x-17\right)\left(\dfrac{1}{15}+\dfrac{1}{14}+\dfrac{1}{13}+\dfrac{1}{12}\right)=0\)
mà \(\dfrac{1}{15}+\dfrac{1}{14}+\dfrac{1}{13}+\dfrac{1}{12}>0\)
nên x-17=0
hay x=17
Vậy: x=17
b) Ta có: \(\dfrac{x+1}{19}+\dfrac{x+2}{18}+\dfrac{x+3}{17}+...+\dfrac{x+18}{2}+18=0\)
\(\Leftrightarrow\dfrac{x+1}{19}+1+\dfrac{x+2}{18}+1+\dfrac{x+3}{17}+1+...+\dfrac{x+18}{2}+1=0\)
\(\Leftrightarrow\dfrac{x+20}{19}+\dfrac{x+20}{18}+\dfrac{x+20}{17}+...+\dfrac{x+20}{2}=0\)
\(\Leftrightarrow\left(x+20\right)\left(\dfrac{1}{19}+\dfrac{1}{18}+\dfrac{1}{17}+...+\dfrac{1}{2}\right)=0\)
mà \(\dfrac{1}{19}+\dfrac{1}{18}+\dfrac{1}{17}+...+\dfrac{1}{2}>0\)
nên x+20=0
hay x=-20
Vậy: x=-20
Giải phương trình:
a) \(\dfrac{x^2}{2}\) + \(\dfrac{18}{x^2}\) = 13(\(\dfrac{x}{2}\) - \(\dfrac{3}{x}\))
b) x(x - 1) + \(\dfrac{1}{x}\)(\(\dfrac{1}{x}\) - 1) = 0
a. ĐKXĐ:...
\(\Leftrightarrow2\left(\dfrac{x^2}{4}+\dfrac{9}{x^2}\right)=13\left(\dfrac{x}{2}-\dfrac{3}{x}\right)\)
\(\Leftrightarrow2\left(\dfrac{x^2}{4}+\dfrac{9}{x^2}-3+3\right)=13\left(\dfrac{x}{2}-\dfrac{3}{x}\right)\)
\(\Leftrightarrow2\left(\dfrac{x}{2}-\dfrac{3}{x}\right)^2+6=13\left(\dfrac{x}{2}-\dfrac{3}{x}\right)\)
Đặt \(\dfrac{x}{2}-\dfrac{3}{x}=t\Rightarrow2t^2-13t+6=0\Rightarrow\left[{}\begin{matrix}t=6\\t=\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{x}{2}-\dfrac{3}{x}=6\\\dfrac{x}{2}-\dfrac{3}{x}=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-12x-6=0\\x^2-x-6=0\end{matrix}\right.\)
\(\Leftrightarrow...\)
b. ĐKXĐ: ...
\(\Leftrightarrow x\left(x-1\right)-\dfrac{x-1}{x^2}=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-\dfrac{1}{x^2}\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3-1\right)=0\)
\(\Leftrightarrow x=1\)
tìm x, biết:
e) \(\dfrac{2}{7-x}\)=\(\dfrac{18}{45}\) f)\(\dfrac{2x+3}{3}\)=\(\dfrac{50}{15}\) g)\(2\dfrac{x}{7}\)=\(\dfrac{75}{35}\) h)\(2\dfrac{3}{x}\)=\(\dfrac{13}{x}\)(x khác 0)
e: =>2/7-x=2/5
=>7-x=5
=>x=2
f: =>2x+3/3=10/3
=>2x+3=10
=>2x=7
=>x=7/2
g: =>(14+x)/7=15/7
=>x+14=15
=>x=1
h: =>(2x+3)/x=13/x
=>2x+3=13
=>2x=10
=>x=5
Tính bằng cách thuận tiện nhất
\(\dfrac{13}{18}\) x \(\dfrac{9}{26}\) x \(\dfrac{2}{3}\) =
\(\dfrac{44}{15}\) x \(\dfrac{5}{22}\) x \(\dfrac{3}{2}\) =
a: =13/26*9/18*2/3=2/3*1/4=2/12=1/6
b: =44/22*5/15*3/2=2*1/3*3/2=2*1/2=1
tính bằng cách thuận tiện nhất a, \(\dfrac{5}{13}\)x\(\dfrac{4}{15}\)x13= b, (\(\dfrac{3}{7}\)+\(\dfrac{5}{2}\))x\(\dfrac{7}{5}\)= c, \(\dfrac{1}{5}\)x\(\dfrac{11}{18}\)+\(\dfrac{11}{18}\)x\(\dfrac{3}{5}\)=
\(a,\dfrac{5}{13}\times\dfrac{4}{15}\times13=\dfrac{5\times4\times13}{13\times5\times3}=\dfrac{4}{3}\\ b,\left(\dfrac{3}{7}+\dfrac{5}{2}\right)\times\dfrac{7}{5}=\dfrac{3}{7}\times\dfrac{7}{5}+\dfrac{5}{2}\times\dfrac{7}{5}=\dfrac{3}{5}+\dfrac{7}{2}=\dfrac{6}{10}+\dfrac{35}{10}=\dfrac{41}{10}\\ c,\dfrac{1}{5}\times\dfrac{11}{18}+\dfrac{11}{18}\times\dfrac{3}{5}=\dfrac{11}{18}\times\left(\dfrac{1}{5}+\dfrac{3}{5}\right)=\dfrac{11}{18}\times\dfrac{4}{5}=\dfrac{22}{45}\)
(11\(\dfrac{7}{18}\) - 9\(\dfrac{13}{18}\)): x - 1\(\dfrac{2}{33}\) : \(\dfrac{7}{11}\) = 1\(\dfrac{2}{3}\)
\(\Leftrightarrow\left(11+\dfrac{7}{18}-9-\dfrac{13}{18}\right):x=\dfrac{5}{3}+\dfrac{35}{33}\cdot\dfrac{11}{7}=\dfrac{10}{3}\)
\(\Leftrightarrow x=\dfrac{5}{3}:\dfrac{10}{3}=\dfrac{1}{2}\)
\(Tìm\) \(x\)∈\(Z\)\(,\) \(biết\)\(:\)
\(a\)) \(\left(x-20\right)+\left(x-19\right)+\left(x-18\right)+...+99+100=100\)
\(b\)) \(213-x.\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2020}}\right):\left(1-\dfrac{1}{2^{2020}}\right)=13\)
a) Quy luật là gì ??
b)
Đặt
\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2020}}\\\Rightarrow2A=1+\dfrac{1}{2}+...+\dfrac{1}{2^{2019}}\\ \Rightarrow2A-A=1-\dfrac{1}{2^{2020}}\Rightarrow A=1-\dfrac{1}{2^{2020}}\)
Suy ra , phương trình trở thành :
213 -x =13
<=> x=200
Rút gọn:
C= \(sin^2\dfrac{\pi}{3}+sin^2\dfrac{5\pi}{6}+sin^2\dfrac{\pi}{9}+sin^2\dfrac{11\pi}{18}+sin^2\dfrac{13\pi}{18}+sin^2\dfrac{2\pi}{9}\)
D=\(cos\left(x-\dfrac{\pi}{3}\right).cos\left(x+\dfrac{\pi}{4}\right)+cos\left(x+\dfrac{\pi}{6}\right).cos\left(x+\dfrac{3\pi}{4}\right)\)
Câu 1: Tìm x, biết:
a)\(x^2-\dfrac{16}{25}=0\) b)\(\dfrac{2}{5}-\left|\dfrac{1}{2}-x\right|=6\)
C2.Tính giá của biểu thức:
a)\(A=1\dfrac{5}{13}-0,25-\left(2\dfrac{5}{9}+\dfrac{18}{13}-\dfrac{1}{4}\right)\)
b)\(\dfrac{\dfrac{3}{5}.7^2-3.5^6+\dfrac{3}{5}.3^9}{\dfrac{3}{4}.7^2-\dfrac{3}{4}.5^7+\dfrac{3}{4}.3^9}\)
a)
x^2-16/25=0
x^2-4^2/5^2=0
=>x-4/5=0
x=0+4/5
x=0/5