Giaỉ các phương trình sau ;
a) 2/x^3-x^2-x+1 = 3/1-x^2 - 1/x+1
b)x+2/x-2 - 2/x^2-2x = 1/x
Giaỉ phương trình sau
\(\sqrt[]{16x}=8\)
mn trình bày cho mik nha , mik ko bt cách trình bày ;((((
\(\sqrt{16x}=8\Leftrightarrow16x=64\Leftrightarrow x=4\)
Ta có: \(\sqrt{16x}=8\)
\(\Leftrightarrow16x=64\)
hay x=4
c: Ta có: \(\sqrt{2x}=\sqrt{5}\)
\(\Leftrightarrow2x=5\)
hay \(x=\dfrac{5}{2}\)
d: Ta có: \(\sqrt{3x-1}=4\)
\(\Leftrightarrow3x-1=16\)
\(\Leftrightarrow3x=17\)
hay \(x=\dfrac{17}{3}\)
Giaỉ phương trình sau:
\(x^4+2x^3-2x^2+2x-3=0\)
Giaỉ phương trình sau :
\(\dfrac{180}{x-4}-\dfrac{180}{x}=\dfrac{1}{2}\)
\(\dfrac{180}{x-4}-\dfrac{180}{x}=\dfrac{1}{2}\)
\(\Leftrightarrow\) \(\dfrac{2x\cdot180}{2x\left(x-4\right)}-\dfrac{2\cdot180\cdot\left(x-4\right)}{2x\left(x-4\right)}=0\)
\(\Leftrightarrow\) \(\dfrac{360x-360x+1440-x^2+4x}{2x\left(x-4\right)}=0\)
\(\Leftrightarrow\) \(\dfrac{-x^2+4x+1440}{2x\left(x-4\right)}=0\)
\(\Leftrightarrow-x^2+4x+1440=0\)
\(\Leftrightarrow-x^2+40x-36x+1440=0\)
\(\Leftrightarrow-x\cdot\left(x-40\right)\cdot\left(-36\right)\cdot\left(x-40\right)=0\)
\(\Leftrightarrow\left(x-40\right)\cdot\left(x-36\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-40=0\\x+36=0\end{matrix}\right.\)
\(x-40=0\)
\(x=0+40\)
\(x=40\)
\(x+36=0\)
\(x=0-36\)
\(x=-36\)
\(\Leftrightarrow\left[{}\begin{matrix}x=40\\x=-36\end{matrix}\right.\)
\(180\left(\dfrac{1}{x-4}-\dfrac{1}{x}\right)=\dfrac{1}{2}\)
\(\dfrac{1}{x-4}-\dfrac{1}{x}=\dfrac{1}{360}\left(đk:x\ne0,4\right)\)
\(\dfrac{x-x+4}{x\left(x-4\right)}=\dfrac{1}{360}\)
\(\dfrac{4}{x\left(x-4\right)}=\dfrac{1}{360}\)
\(x^2-4x=1440\)
\(x^2-4x+4=1444\)
\(\left(x-2\right)^2=1444=38^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=38\\x-2=-38\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=40\\x=-36\end{matrix}\right.\)
Giaỉ các bất phương trình sau rồi biểu diễn trên trục số
a) x+17<10 b) 9-2x<0 c) -11-3x≥0
d) -5x≤18+x
GIÚP MIK NHA MN
a: Ta có: x+17<10
nên x<-7
b: Ta có: 9-2x<0
\(\Leftrightarrow2x>9\)
hay \(x>\dfrac{9}{2}\)
c: Ta có: \(-3x-11\ge0\)
\(\Leftrightarrow-3x\ge11\)
hay \(x\le-\dfrac{11}{3}\)
Giaỉ các phương trình sau
a) 3/x^2+x-2 - 1/x-1 = -7/x+2
b) 2/-x^2+6x-8 - x-1/x-2 = x+3/x-4
a: Ta có: \(\dfrac{3}{x^2+x-2}-\dfrac{1}{x-1}=\dfrac{-7}{x+2}\)
\(\Leftrightarrow3-\left(x+2\right)=-7\left(x-1\right)\)
\(\Leftrightarrow3-x-2+7x-7=0\)
\(\Leftrightarrow6x-6=0\)
hay x=1(loại
b: Ta có: \(\dfrac{2}{-x^2+6x-8}-\dfrac{x-1}{x-2}=\dfrac{x+3}{x-4}\)
\(\Leftrightarrow\dfrac{-2}{\left(x-2\right)\left(x-4\right)}-\dfrac{\left(x-1\right)\left(x-4\right)}{\left(x-2\right)\left(x-4\right)}=\dfrac{\left(x+3\right)\left(x-2\right)}{\left(x-4\right)\left(x-2\right)}\)
Suy ra: \(-2-x^2+5x-4=x^2+x-6\)
\(\Leftrightarrow-x^2+5x-6-x^2-x+6=0\)
\(\Leftrightarrow-2x^2+4x=0\)
\(\Leftrightarrow-2x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=2\left(loại\right)\end{matrix}\right.\)
\(\dfrac{3}{x^2+x-2}-\dfrac{1}{x-1}=-\dfrac{7}{x+2}\)
\(\Rightarrow\dfrac{3}{\left(x^2-x\right)+\left(2x-2\right)}-\dfrac{1}{x-1}=-\dfrac{7}{x+2}\)
\(\Rightarrow\dfrac{3}{x\left(x-1\right)+2\left(x-1\right)}-\dfrac{1}{x-1}=-\dfrac{7}{x+2}\)
\(\Rightarrow\dfrac{3}{\left(x+2\right)\left(x-1\right)}-\dfrac{1}{x-1}+\dfrac{7}{x+2}=0\)
\(\Rightarrow\dfrac{3}{\left(x+2\right)\left(x-1\right)}-\dfrac{x+2}{\left(x+2\right)\left(x-1\right)}+\dfrac{7\left(x-1\right)}{\left(x+2\right)\left(x-1\right)}=0\)
\(\Rightarrow\dfrac{3-\left(x+2\right)+7\left(x-1\right)}{\left(x+2\right)\left(x-1\right)}=0\)
\(\Rightarrow3-x-2+7x-7=0\)
\(\Rightarrow6x-6=0\)
\(\Rightarrow x=1\)
Giaỉ các phương trình sau
a) x-3/x-2 -x-2/x-4 = 5/21
b)x+1/x-2 - x-1/x+2 = 2(x^2+2)/x^2-4
b: Ta có: \(\dfrac{x+1}{x-2}-\dfrac{x-1}{x+2}=\dfrac{2\left(x^2+2\right)}{x^2-4}\)
\(\Leftrightarrow x^2+3x+2-x^2+3x-2-2x^2-4=0\)
\(\Leftrightarrow-2x^2+6x-4=0\)
a=-2; b=6; c=-4
Vì a+b+c=0 nên phương trình có hai nghiệm phân biệt là:
\(x_1=1\left(nhận\right);x_2=\dfrac{c}{a}=2\left(loại\right)\)
giaỉ phương trình sau: (12x-1)(6x-1)(4x-1)(3x-1)=5
(12x-1)(6x-1)(4x-1)(3x-1)=5
<=>(12x-1)(12x-2)(12x-3)(12x-4)=40
<=>[(12x-1)(12x-4)] [(12x-2)(12x-3)] =40
<=>(144x^2 - 60x + 4) (144x^2 - 60x + 6) =40
đặt 144x^2 - 60x +4 = t =>144x^2 - 60x +6 = t+2
ta có phương trình:
t ( t+2 ) =40
<=> t^2 + 2t -40 =0
<=> (t+1)^2 -39 =0
<=> t+1=\(\sqrt{39}\) hoặc t+1=\(-\sqrt{39}\) <=> x=\(\sqrt{39}\) -1 hoặc x=\(-\sqrt{39}\) -1
Giaỉ phương trình sau ;
4/x^2+2x-3 = 2x-5/x+3 - 2x/x-1
Ta có: \(\dfrac{4}{x^2+2x-3}=\dfrac{2x-5}{x+3}-\dfrac{2x}{x-1}\)
\(\Leftrightarrow\dfrac{\left(2x-5\right)\left(x-1\right)-2x\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}=\dfrac{4}{\left(x+3\right)\left(x-1\right)}\)
Suy ra: \(2x^2-2x-5x+5-2x^2-6x=4\)
\(\Leftrightarrow13x=-1\)
hay \(x=-\dfrac{1}{13}\)
1 Giaỉ các phương trình sau ;
a)\(\sqrt[]{4x^2-4x+9=3}\) b) \(\sqrt[]{16x=8}\)
c) \(\sqrt{2x}=\sqrt{5}\) d) \(\sqrt[]{3x-1}=4\)
e)\(\sqrt[]{4\left(1-x\right)^2}-6=0\) g) \(\sqrt[]{x^2-x+16=4}\)
mn giúp e nha , e đang cần gấp ^^
c: Ta có: \(\sqrt{2x}=\sqrt{5}\)
\(\Leftrightarrow2x=5\)
hay \(x=\dfrac{5}{2}\)
d: Ta có: \(\sqrt{3x-1}=4\)
\(\Leftrightarrow3x-1=16\)
\(\Leftrightarrow3x=17\)
hay \(x=\dfrac{17}{3}\)
Ta có: \(\sqrt{4\cdot\left(1-x\right)^2}=6\)
\(\Leftrightarrow2\left|x-1\right|=6\)
\(\Leftrightarrow\left|x-1\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=3\\x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\)
Ta có: \(\sqrt{4x^2-4x+9}=3\)
\(\Leftrightarrow4x^2-4x=0\)
\(\Leftrightarrow4x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)