giúp mk vs ạ
cần gấp ạ
căn(x+1) +2căn(2x+3)>=2x+2
K
Khách
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17 tháng 8 2021
\(a)\)
\(\frac{1}{x+1}-\frac{x-1}{x}=\frac{3x+1}{x\left(x+1\right)}\)
\(\Leftrightarrow x-x^2+1=3x+1\)
\(\Leftrightarrow x^2-2x=0\)
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
\(b)\)
\(\frac{\left(x+2\right)^2}{2x-3}-\frac{1}{1}=\frac{x^2+10}{2x-3}\)
\(\Leftrightarrow x^2+4x+4-2x-3=x^2+10\)
\(\Leftrightarrow x^2+2x+1=x^2+10\)
\(\Leftrightarrow2x-9=0\)
\(\Leftrightarrow2x=9\)
\(\Leftrightarrow x=\frac{2}{9}\)
T
1
20 tháng 8 2021
\(1,\)
\(2x\left(x-3\right)-\left(3-x\right)=0\)
\(\Leftrightarrow2x\left(x-3\right)+\left(x-3\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\x-3=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=3\end{cases}}\)
\(2,\)
\(3x\left(x+5\right)-6\left(x+5\right)=0\)
\(\Leftrightarrow\left(3x-6\right)\left(x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-6=0\\x+5=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-5\end{cases}}\)
\(3,\)
\(x^4-x^2=0\)
\(\Leftrightarrow x^2\left(x^2-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\x^2-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)
\(4,\)
\(x^2-2x=0\)
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
\(5,\)
\(x\left(x+6\right)-10\left(x-6\right)=0\)
\(\Leftrightarrow x^2+6x-10x+60=0\)
\(\Leftrightarrow x^2-4x+60=0\)
\(\Leftrightarrow x^2-4x+4+56=0\)
\(\Leftrightarrow\left(x-2\right)^2=-56\)(Vô lý)
=> Phương trình vô nghiệm
TL
1 tháng 7 2021
Bằng 0 chứ nhỉ em ?
(x-5) . (2x-4)= 0
\(\left[{}\begin{matrix}x-5=0\\2x-4=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=5\\2x=4\end{matrix}\right.< =>\left[{}\begin{matrix}x=5\\x=2\end{matrix}\right.\)
TL
1 tháng 7 2021
(x-5) . (2x-4)= 5
<=> 2x^2 - 4x - 10x + 20 = 5
<=> 2x^2 - 14x + 15 = 0
Giải pt bâc hai ra đc : \(x=\dfrac{7+\sqrt{19}}{2}\)và \(x2=\dfrac{7-\sqrt{19}}{2}\)
LT
5 tháng 8 2020
Bài 9 : Tìm x, biết :
a, (x - 2)(x - 3) + (x - 2) - 1 = 0
\(\Leftrightarrow\left(x-2\right)\left(x-3+1\right)-1=0\)
\(\Leftrightarrow\left(x-2\right)^2-1=0\)
\(\Leftrightarrow\left(x-2+1\right)\left(x-2-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
Vậy x ={1; 3}
b, (x + 2)2 - 2x(2x + 3) = (x + 1)2
\(\Leftrightarrow\left(x+2\right)^2-\left(x+1\right)^2-2x\left(2x+3\right)=0\)
\(\Leftrightarrow\left(x+2+x+1\right)\left(x+2-x-1\right)-2x\left(2x+3\right)=0\)
\(\Leftrightarrow2x+3-2x\left(2x+3\right)=0\)
\(\Leftrightarrow\left(2x+3\right)\left(1-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\1-2x=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{3}{2}\\x=\frac{1}{2}\end{matrix}\right.\)
Vậy \(x=\left\{-\frac{3}{2};\frac{1}{2}\right\}\)
c, 6x3 + x2 = 2x
\(\Leftrightarrow6x^3+x^2-2x=0\)
\(\Leftrightarrow x\left(6x^2+x-2\right)=0\)
\(\Leftrightarrow x\left(6x^2+4x-3x-2\right)=0\)
\(\Leftrightarrow x\left[2x\left(3x+2\right)-\left(3x+2\right)\right]=0\)
\(\Leftrightarrow x\left(3x+2\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\3x+2=0\\2x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\frac{2}{3}\\x=\frac{1}{2}\end{matrix}\right.\)
Vậy \(x=\left\{0;-\frac{2}{3};\frac{1}{2}\right\}\)
CH
Cô Hoàng Huyền
Admin
VIP
8 tháng 7 2016
a, \(4^x-10.2^x+16=0\Leftrightarrow\left(2^x\right)^2-10.2^x+16=0\)
Đặt \(2^x=t\Rightarrow t^2-10t+16=0\Leftrightarrow\orbr{\begin{cases}t=8\\t=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}\)
b. Đặt \(2x^2-3x-1=t\Rightarrow t^2-3\left(t-4\right)-16=0\)
\(\Leftrightarrow t^2-3t-28=0\Leftrightarrow\orbr{\begin{cases}t=7\\t=-4\end{cases}}\)
Thế vào rồi giải tiếp em nhé.
\(\sqrt{x+1}+2\sqrt{2x+3}\ge2x+2\)
\(\Leftrightarrow\sqrt{x+1}-2+2\sqrt{2x+3}-6-2x+6\ge0\)
\(\Leftrightarrow\frac{x+1-4}{\sqrt{x+1}+2}+\frac{2\cdot\left(2x+3-9\right)}{\sqrt{2x+3}+3}-2\left(x-3\right)\ge0\)
\(\Leftrightarrow\frac{x-3}{\sqrt{x+1}+2}+\frac{4\cdot\left(x-3\right)}{\sqrt{2x+3}+3}-2\left(x-3\right)\ge0\)
\(\Leftrightarrow\left(x-3\right)\cdot\left(\frac{1}{\sqrt{x+1}+2}+\frac{4}{\sqrt{2x+3}+3}-2\right)\ge0\)
Xét \(\frac{1}{\sqrt{x+1}+2}+\frac{4}{\sqrt{2x+3}+3}-2=\frac{\sqrt{x+\frac{3}{2}}\cdot\sqrt{x+1}+\sqrt{\frac{x+1}{2}}+\frac{3}{2}\sqrt{x+\frac{3}{2}}+\frac{1}{2\sqrt{2}}}{\left(\sqrt{x+1}+2\right)\left(\sqrt{2x+3}+3\right)}\ge0\)
Do đó \(x-3\ge0\Leftrightarrow x\ge3\)
Vậy...