bài 1 : tìm x
a) x+ ( x+1 ) + x+2 ) + ... + ( x+ 2003) = 2004
b) \(2\frac{1}{3}\) + ( x - \(\frac{3}{2}\) ) = \(3-\frac{3}{2}\) . x
Làm nhanh giúp mình đi mình đang cần gấp
tìm x biết
a. \(\frac{1}{4}.\left\{3-\frac{1}{2}\left[1+\frac{1}{2}\left(\sqrt{2x+1}-\frac{1}{2}\right)\right]\right\}=2\)
b. \(\sqrt{1+2+3+...+\left(x-1\right)+x+\left(x-1\right)+...+3+2+1}=2010\)
giúp mình nha mình đang cần gấp
tìm số nguyên x
a) (x-3)+(x-2)+(x-1)+....+10+11=11
b) \(\frac{2}{3}\).\(\left[\frac{1}{2}+\frac{3}{4}-\frac{1}{3}\right]\)<= x <=\(4\frac{1}{3}.\left[\frac{1}{2}-\frac{1}{6}\right]\)
giúp mình nha mai mình đi học rồi. minh cần gấp lắm bạn nào giải được thì mau giải giúp mình nha
Tìm x biết:
\(\left(\frac{1}{x}-\frac{2}{3}\right)^2-\frac{1}{16}=0\)
mình đang cần gấp , ai nhanh mình tick
\(\left(\frac{1}{x}-\frac{2}{3}\right)^2-\frac{1}{16}=0\)
\(\Leftrightarrow\left(\frac{1}{x}-\frac{2}{3}\right)^2-\left(\frac{1}{4}\right)^2=0\)
\(\Leftrightarrow\left(\frac{1}{x}-\frac{2}{3}+\frac{1}{4}\right)\left(\frac{1}{x}-\frac{2}{3}-\frac{1}{4}\right)=0\)
\(\Leftrightarrow\left(\frac{1}{x}-\frac{5}{12}\right)\left(\frac{1}{x}-\frac{11}{12}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{x}-\frac{5}{12}=0\\\frac{1}{x}-\frac{11}{12}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{x}=\frac{5}{12}\\\frac{1}{x}=\frac{11}{12}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{12}{11}\\x=\frac{12}{5}\end{cases}}\)
Vậy....
\(\left(\frac{1}{x}-\frac{2}{3}\right)^2-\frac{1}{16}=0\)
\(\Rightarrow\left(\frac{1}{x}-\frac{2}{3}\right)^2=\frac{1}{16}\)
\(\Rightarrow\left(\frac{1}{x}-\frac{2}{3}\right)^2=\left(\frac{1}{4}\right)^2\)
\(\Rightarrow\frac{1}{x}-\frac{2}{3}=\frac{1}{4}\)
\(\Rightarrow\frac{1}{x}=\frac{11}{12}\)
\(\Rightarrow x=\frac{11}{12}\)
\(=>\left(\frac{1}{x}-\frac{2}{4}\right)^2=\frac{1}{16}\)
=> \(\orbr{\begin{cases}\frac{1}{x}-\frac{2}{4}=\frac{1}{4}\\\frac{1}{x}-\frac{2}{4}=-\frac{1}{4}\end{cases}}\)
=>\(\orbr{\begin{cases}\frac{1}{x}=\frac{3}{4}\\\frac{1}{x}=\frac{1}{4}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{4}{3}\\x=4\end{cases}}\)
học tốt
a)\(\frac{x^2+5x}{5x^2+x^3}\) b) \(\frac{x^4+x^2+1}{x^3+1}\)
giúp mình với mình đang cần gấp
a) \(\frac{x^2+5x}{5x^2+x^3}\)
\(=\frac{x\left(x+5\right)}{x^2\left(x+5\right)}=\frac{1}{x}\)
b) \(\frac{x^4+x^2+1}{x^3+1}\)
\(=\frac{\left(x^2+x+1\right)\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\frac{x^2+x+1}{x+1}\)
\(a)\frac{x^2+5x}{5x^2+x^3}=\frac{x\left(x+5\right)}{x^2\left(5+x\right)}=\frac{1}{x}\)
\(a,\frac{x^2+5x}{5x^2+x^3}=\frac{x\left(x+5\right)}{x^2\left(x+5\right)}=\frac{1}{x}\)
\(b,\frac{x^4+x^2+1}{x^3+1}=\frac{x^4+x^3+x^2-\left(x^3-1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{x^2\left(x^2+x+1\right)-\left(x-1\right)\left(x^2+x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\frac{\left(x^2+x+1\right)\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{x^2+x+1}{x+1}\)
hok tốt nha
\(a)\frac{x-1}{6}=\frac{2x+3}{7}\)
\(b)(2x^2-\frac{1}{2}x).(x^2+1)=0\)
\(c)\frac{4}{3}-\frac{1}{3}|x-1|=\frac{1}{2}\)
làm giúp mình nha!!!!
mình đang cần gấp!!!
a) \(\frac{x-1}{6}=\frac{2x+3}{7}\)
\(\Leftrightarrow7\left(x-1\right)=6\left(2x+3\right)\)
\(\Leftrightarrow7x-7=12x+18\)
\(\Leftrightarrow5x+18=-7\)
\(\Leftrightarrow5x=-25\)
\(\Leftrightarrow x=-5\)
b) \(\left(2x^2-\frac{1}{2}x\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow x\left(2x-\frac{1}{2}\right)\left(x^2+1\right)=0\)
Vì \(x^2+1>0\)nên \(\orbr{\begin{cases}x=0\\2x-\frac{1}{2}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{4}\end{cases}}\)
c) \(\frac{4}{3}-\frac{1}{3}\left|x-1\right|=\frac{1}{2}\)
=> \(\frac{1}{3}\left|x-1\right|=\frac{4}{3}-\frac{1}{2}\)
=> \(\frac{1}{3}\left|x-1\right|=\frac{5}{6}\)
=> \(\left|x-1\right|=\frac{5}{6}:\frac{1}{3}\)
=> \(\left|x-1\right|=\frac{5}{2}\)
=> \(\orbr{\begin{cases}x-1=\frac{5}{2}\\x-1=-\frac{5}{2}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{3}{2}\end{cases}}\)
Tìm các số nguyên x, biết: \(-4\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{6}\right)\le x\le\frac{-2}{3}.\left(\frac{1}{3}-\frac{1}{2}-\frac{3}{4}\right)\)
Mình đag cần rất gấp. Ai làm nhanh mình tick. Mong mn giúp mình với
Tìm các số nguyên x biết : \(-4\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{6}\right)\le x\le-\frac{2}{3}.\left(\frac{1}{3}-\frac{1}{2}-\frac{3}{4}\right)\)
Mình đag cần rất gấp. Ai lm nhanh mình tick. Mong mn giúp mình với
\(-4\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{6}\right)\le x\le-\frac{2}{3}.\left(\frac{1}{3}-\frac{1}{2}-\frac{3}{4}\right)\)
\(\Rightarrow-\frac{13}{3}.\left(\frac{3}{6}-\frac{1}{6}\right)\le x\le-\frac{2}{3}.\left(\frac{4}{12}-\frac{6}{12}-\frac{9}{12}\right)\)
\(\Rightarrow-\frac{13}{3}.\frac{2}{6}\le x\le-\frac{2}{3}.\frac{-11}{12}\)
\(\Rightarrow\frac{-13}{9}\le x\le\frac{11}{18}\)
\(\Rightarrow\frac{-26}{18}\le x\le\frac{11}{18}\)
=> -1,44444444444........... ≤ x ≤ 0,6111111111...........
Mà x ∈ Z
=> x ∈ { -1 ; 0 }
\(x\in\varnothing\)
tìm x biết:
\(\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)\times x=\frac{2103}{1}+\frac{2012}{2}+...+\frac{2}{2012}+\frac{1}{2013}\)
Giúp mình với mình đang cần gấp
giúp mình với ạ , mình đang cần gấp !!!
a,\(\hept{\begin{cases}3\left(x+1\right)+2\left(x+2y\right)=4\\4\left(x+1\right)-\left(x+2y\right)=9\end{cases}}\)
b, \(\hept{\begin{cases}x+\frac{1}{y}=\frac{-1}{2}\\2x-\frac{3}{y}=\frac{-7}{2}\end{cases}}\)
c,\(\hept{\begin{cases}\frac{x+2}{x+1}+\frac{2}{y-2}=6\\\frac{5}{x+1}-\frac{1}{y-2}=3\end{cases}}\)
c) Ta có: \(\left\{{}\begin{matrix}\dfrac{x+2}{x+1}+\dfrac{2}{y-2}=6\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x+1}+\dfrac{2}{y-2}=5\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{x+1}+\dfrac{10}{y-2}=25\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{11}{y-2}=22\\\dfrac{1}{x+1}+\dfrac{2}{y-2}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y-2=\dfrac{1}{2}\\\dfrac{1}{x+1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+1=1\\y-2=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=\dfrac{5}{2}\end{matrix}\right.\)