Tìm x biết: \(\left|\dfrac{3}{x-2}\right|+\dfrac{3}{x-2}=0\)
Các bạn giúp mik nhanh vs ạ
Rút gọn các biểu thức sau:
a/\(\left(x+\dfrac{1}{3}x+\dfrac{1}{9}\right)\left(x-\dfrac{1}{3}\right)-\left(x-\dfrac{1}{3^{ }}\right)^2\)
b/\(\left(x_{ }^2-2\right)^3-x\left(x+1\right)\left(x-1\right)+x\left(x-3\right)\)
MẤY BẠN GIÚP MK VS Ạ AI NHANH MK VOTE NHA
a) \(=x^3-\dfrac{1}{27}-x^2+\dfrac{2}{3}x-\dfrac{1}{9}=x^3-x^2+\dfrac{2}{3}x-\dfrac{2}{27}\)
b) \(=x^6-6x^4+12x^2-8-x^3+x+x^2-3x=x^6-6x^4-x^3+13x^2-2x-8\)
RÚT GỌN BIỂU THỨC SAU
\(\left(x+\dfrac{1}{3}x+\dfrac{1}{9}\right)\left(x-\dfrac{1}{3}\right)-\left(x-\dfrac{1}{3}\right)^2\)
MẤY BẠN GIÚP MK VS Ạ AI NHANH MK VOTE NHA
\(=\left(x-\dfrac{1}{3}\right)\left(\dfrac{4}{3}x+\dfrac{1}{9}-x+\dfrac{1}{3}\right)\\ =\left(x-\dfrac{1}{3}\right)\left(\dfrac{1}{3}x+\dfrac{4}{9}\right)\\ =\dfrac{1}{3}x^2+\dfrac{4}{9}x-\dfrac{1}{9}x-\dfrac{4}{27}\\ =\dfrac{1}{3}x^2+\dfrac{1}{3}x-\dfrac{4}{27}\)
tìm nghiệm của đa thức sau
a,\(3x-\dfrac{2}{5}\)
b,\(\left(x-3\right)\).\(\left(2x+8\right)\)
c, \(3.x^2\)-\(x\)-\(4\)
mn giúp mik vs ạ , mik c.on trc ạ
a)\(3x-\dfrac{2}{5}=0=>3x=\dfrac{2}{5}=>x=\dfrac{2}{15}\)
b)\(\left(x-3\right)\left(2x+8\right)=0=>\left[{}\begin{matrix}x-3=0\\2x=-8\end{matrix}\right.=>\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\)
c)\(3x^2-x-4=0=>3x^2+3x-4x-4=0=>\left(3x-4\right)\left(x+1\right)=0\)
\(=>\left[{}\begin{matrix}3x=4\\x+1=0\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-1\end{matrix}\right.\)
tìm x biết
\(\dfrac{7}{9}:\left(2+\dfrac{3}{4}.x\right)+\dfrac{5}{9}=\dfrac{23}{27}\)
|x|\(-\dfrac{3}{4}=\dfrac{5}{3}\)
\(\left|2.x-\dfrac{1}{3}\right|+\dfrac{5}{6}=1\)
giúp mk vs nhanh lên mình đang bận
b) Ta có: \(\left|x\right|-\dfrac{3}{4}=\dfrac{5}{3}\)
\(\Leftrightarrow\left|x\right|=\dfrac{5}{3}+\dfrac{3}{4}=\dfrac{20}{12}+\dfrac{9}{12}=\dfrac{29}{12}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{29}{12}\\x=-\dfrac{29}{12}\end{matrix}\right.\)
c) Ta có: \(\left|2x-\dfrac{1}{3}\right|+\dfrac{5}{6}=1\)
\(\Leftrightarrow\left|2x-\dfrac{1}{3}\right|=\dfrac{1}{6}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{1}{3}=\dfrac{1}{6}\\2x-\dfrac{1}{3}=\dfrac{-1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{1}{6}+\dfrac{1}{3}=\dfrac{1}{2}\\2x=\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=\dfrac{1}{12}\end{matrix}\right.\)
\(\left(\dfrac{2}{3}x+\dfrac{1}{2}\right).\left(-2x+3\right)=0\)
giúp mik với ạ
`(2/3 x +1/2) (-2x+3)=0`
\(\Rightarrow\left[{}\begin{matrix}\dfrac{2}{3}x+\dfrac{1}{2}=0\\-2x+3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{2}{3}x=-\dfrac{1}{2}\\-2x=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}.\dfrac{3}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{4}\\x=\dfrac{3}{2}\end{matrix}\right.\)
\(\left(\dfrac{2}{3}x+\dfrac{1}{2}\right)\cdot\left(-2x+3\right)=0\\ =>\left[{}\begin{matrix}\dfrac{2}{3}x+\dfrac{1}{2}=0\\-2x+3=0\end{matrix}\right.\\ =>\left[{}\begin{matrix}\dfrac{2}{3}x=-\dfrac{1}{2}\\-2x=-3\end{matrix}\right.\\ =>\left[{}\begin{matrix}x=-\dfrac{3}{4}\\x=\dfrac{3}{2}\end{matrix}\right.\)
tính nhanh
\(\left(1-\dfrac{1}{2}\right)x\left(1-\dfrac{1}{3}\right)x\left(1-\dfrac{1}{4}\right)x\left(1-\dfrac{1}{5}\right)=\)
làm ơn giúp mik với ạ ^^
\(\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times\dfrac{4}{5}=\dfrac{1}{5}\)
\(\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{4}\right)\times\left(1-\dfrac{1}{5}\right)\)
\(=\left(\dfrac{2}{2}-\dfrac{1}{2}\right)\times\left(\dfrac{3}{3}-\dfrac{1}{3}\right)\times\left(\dfrac{4}{4}-\dfrac{1}{4}\right)\times\left(\dfrac{5}{5}-\dfrac{1}{5}\right)\)
\(=\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times\dfrac{4}{5}\)
\(=\dfrac{1\times2\times3\times4}{2\times3\times4\times5}\)
\(=\dfrac{1}{5}\)
\(\left(x^2-1\dfrac{9}{16}\right).\left(x^3+\dfrac{1}{8}\right)=0\)
giúp mik với ạ
Mn ơi giúp mik câu này với ạ !
cho biểu thức P=\(\left(3-\dfrac{3}{\sqrt{x}-1}\right)\):\(\left(\dfrac{x+2}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}}{\sqrt{x}+2}\right)\) với x ≥ 0 ;x ≠ 4
a) Rút gọn biểu thức P
b) Tìm giá trị của x để P=\(\dfrac{4\sqrt{x}-1}{\sqrt{x}}\)
a) \(P=\left(3-\dfrac{3}{\sqrt{x}-1}\right):\left(\dfrac{x+2}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}}{\sqrt{x}+2}\right)\)
\(=\left(\dfrac{3\left(\sqrt{x}-1\right)-3}{\sqrt{x}-1}\right):\left[\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x+2}\right)}-\dfrac{\sqrt{x}}{\sqrt{x}+2}\right]\)
\(=\dfrac{3\sqrt{x}-3-3}{\sqrt{x}-1}:\dfrac{x+2-\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{3\sqrt{x}-6}{\sqrt{x}-1}:\dfrac{x+2-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{3\sqrt{x}-6}{\sqrt{x}-1}:\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{3\sqrt{x}-6}{\sqrt{x}-1}:\dfrac{1}{\sqrt{x}-1}\)
\(=\dfrac{3\sqrt{x}-6}{\sqrt{x}-1}.\left(\sqrt{x}-1\right)\)
\(=3\sqrt{x}-6\)
b) \(P=\dfrac{4\sqrt{x}-1}{\sqrt{x}}\)
\(\Leftrightarrow3\sqrt{x}-6=\dfrac{4\sqrt{x}-1}{\sqrt{x}}\) (1)
ĐKXĐ: \(x>0\)
\(\left(1\right)\Leftrightarrow3x-6\sqrt{x}=4\sqrt{x}-1\)
\(\Leftrightarrow3x-6\sqrt{x}-4\sqrt{x}+1=0\)
\(\Leftrightarrow3x-10\sqrt{x}+1=0\) (2)
Đặt \(t=\sqrt{x}\ge0\)
\(\left(2\right)\Leftrightarrow3t^2-10t+1=0\)
\(\Delta'=25-4=22\)
Phương trình có hai nghiệm phân biệt:
\(t_1=\dfrac{5+\sqrt{22}}{3}\) (nhận)
\(t_2=\dfrac{5-\sqrt{22}}{3}\) (nhận)
Với \(t=\dfrac{5+\sqrt{22}}{3}\) \(\Leftrightarrow\sqrt{x}=\dfrac{5+\sqrt{22}}{3}\Leftrightarrow x=\dfrac{47+10\sqrt{22}}{9}\) (nhận)
Với \(t=\dfrac{5-\sqrt{22}}{3}\Leftrightarrow\sqrt{x}=\dfrac{5-\sqrt{22}}{3}\Leftrightarrow x=\dfrac{47-10\sqrt{22}}{9}\) (nhận)
Vậy \(x=\dfrac{47+10\sqrt{22}}{9};x=\dfrac{47-10\sqrt{22}}{9}\) thì \(P=\dfrac{4\sqrt{x}-1}{\sqrt{x}}\)
a: \(P=\dfrac{3\sqrt{x}-3-3}{\sqrt{x}-1}:\dfrac{x+2-x+\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{3\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+2}=3\sqrt{x}-6\)
b: P=(4căn x-1)/căn x
=>3x-6căn x-4căn x+1=0
=>3x-10căn x+1=0
=>x=(47+10căn 22)/9 hoặc x=(47-10căn 22)/9
giúp mik vs ạ
A=\(\dfrac{x^2+x+1}{1-x^3}+\left(\dfrac{x}{x^2-2x+1}+\dfrac{x}{1-x^2}\right)\cdot\dfrac{x^2-1}{x^2+1}\)
với x≠ 1,x≠-1
a) rút gọn biểu thức A
b) Tìm tất cả các giá trị của x để biểu thức A2 nhận giá trị nguyên