hãy giải pt sau:
(2x-1)2=9
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8 tháng 5 2017
Đặt t=x2-2x+3(t\(\ge\)2)
PTTT: \(\dfrac{1}{t-1}+\dfrac{1}{t}=\dfrac{9}{2\left(t+1\right)}\)
<=>2t2+2t+2t2-2=9t2-9
<=>5t2-2t-7=0
<=>(t+1)(5t-7)=0
Do t\(\ge\)2
=>t+1>0 5t-7>0
Vậy pt vô nghiệm
HN
9 tháng 5 2017
\(\dfrac{1}{x^2-2x+2}+\dfrac{1}{x^2-2x+3}=\dfrac{9}{2\left(x^2-2x+4\right)}\)
Đặt \(t=x^2-2x+2=\left(x-1\right)^2+1\ge1\)
Thì ta có:
\(PT\Leftrightarrow\dfrac{1}{t}+\dfrac{1}{t+1}=\dfrac{9}{2\left(t+2\right)}\)
\(\Leftrightarrow5t^2-t-4=0\)
\(\Leftrightarrow\left(5t^2-5t\right)+\left(4t-4\right)=0\)
\(\Leftrightarrow\left(t-1\right)\left(5t+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5t+4=0\\t-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}t=-\dfrac{4}{5}\left(l\right)\\t=1\end{matrix}\right.\)
\(\Rightarrow x^2-2x+2=1\)
\(\Leftrightarrow x=1\)
Vậy PT có 1 nghiệm là x = 1
16 tháng 7 2019
\(2x^2+3x+3=5\sqrt{2x^2+3x+9}\)
\(\Leftrightarrow2x^2+3x+3=5\sqrt{2x^2+3x+3+6}\)(*)
Đặt \(2x^2+3x+3=a\)
(*) \(\Leftrightarrow a=5\sqrt{a+6}\)
\(\Leftrightarrow a^2=25\left(a+6\right)\)
\(\Leftrightarrow a^2-25a-150=0\)
\(\Leftrightarrow\left(a-30\right)\left(a+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=30\\a=-5\end{matrix}\right.\)
Trả lại biến cũ: \(2x^2+3x+3=30\Leftrightarrow2x^2+3x-27=0\)\(\Leftrightarrow\left(x-3\right)\left(2x+9\right)=0\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\frac{9}{2}\end{matrix}\right.\)
\(2x^2+3x+3=-5\Leftrightarrow2x^2+3x+8=0\)\(\Leftrightarrow\left(x\sqrt{2}+\frac{3\sqrt{2}}{4}\right)^2=-\frac{55}{8}\left(L\right)\)
S
16 tháng 7 2019
Đat: \(2x^2+3x+3=a\)
\(\Rightarrow a=5\sqrt{a+6}\Leftrightarrow a^2=25a+150\Leftrightarrow a^2-25a-150=0\Leftrightarrow\left(a-12,5\right)^2=6,25\Leftrightarrow\left[{}\begin{matrix}a=10\\a=15\end{matrix}\right.\) \(+,a=10\Leftrightarrow x^2+3x+3=10\Leftrightarrow\left(x+\frac{3}{2}\right)^2=9,25\Leftrightarrow x=\pm\sqrt{9,25}-\frac{3}{2}\)
\(+,a=15\Leftrightarrow x^2+3x+2,25=14,25\Leftrightarrow\left(x+\frac{3}{2}\right)^2=14,25\Leftrightarrow x=\pm\sqrt{14,25}-\frac{3}{2}\)
TL
1
12 tháng 5 2023
c: =>(x+2)(x+3)(x-5)(x-6)=180
=>(x^2-3x-10)(x^2-3x-18)=180
=>(x^2-3x)^2-28(x^2-3x)=0
=>x(x-3)(x-7)(x+4)=0
=>\(x\in\left\{0;3;7;-4\right\}\)
c: =>(x-3)(x+2)(2x+1)(3x-1)=0
=>\(x\in\left\{3;-2;-\dfrac{1}{2};\dfrac{1}{3}\right\}\)
BT
7
DL
10 tháng 6 2017
\(7-\left(2x+4\right)=-\left(x+4\right)\)
\(\Rightarrow7-\left(2x+4\right)=-\left(2x-3\right)\)
\(\Rightarrow-\left(2x-3\right)=-\left(x+4\right)\)
\(\Rightarrow3-2x=-x-4\)
\(\Leftrightarrow-x=-7\)
\(\Leftrightarrow x=7\)
DT
4
22 tháng 4 2021
a) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
Ta có: \(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)
\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{12}{\left(x-2\right)\left(x+2\right)}+\dfrac{x^2-4}{\left(x-2\right)\left(x+2\right)}\)
Suy ra: \(x^2+3x+2-5x+10=12+x^2-4\)
\(\Leftrightarrow x^2-2x+12-8-x^2=0\)
\(\Leftrightarrow-2x+4=0\)
\(\Leftrightarrow-2x=-4\)
hay x=2(loại)
Vậy: \(S=\varnothing\)
22 tháng 4 2021
b) Ta có: \(\left|2x+6\right|-x=3\)
\(\Leftrightarrow\left|2x+6\right|=x+3\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+6=x+3\left(x\ge-3\right)\\-2x-6=x+3\left(x< -3\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-x=3-6\\-2x-x=3+6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(nhận\right)\\x=-3\left(loại\right)\end{matrix}\right.\)
Vậy: S={-3}
HT
1
30 tháng 5 2022
\(ĐK:x\in R\)
\(\sqrt{x^2+x+4}+\sqrt{x^2+x+1}=\sqrt{2x^2+2x+9}\) (*)
Đặt \(x^2+x+1=a;a\ge0\)
\(\rightarrow\left\{{}\begin{matrix}x^2+x+4=a+3\\2x^2+2x+9=2a+7\end{matrix}\right.\)
(*) \(\Rightarrow\sqrt{a+3}+\sqrt{a}=\sqrt{2a+7}\)
\(\Leftrightarrow\left(\sqrt{a+3}+\sqrt{a}\right)^2=\left(\sqrt{2a+7}\right)^2\)
\(\Leftrightarrow a+3+a+2\sqrt{a\left(a+3\right)}=2a+7\)
\(\Leftrightarrow2\sqrt{a\left(a+3\right)}=4\)
\(\Leftrightarrow\sqrt{a\left(a+3\right)}=2\)
\(\Leftrightarrow a\left(a+3\right)=4\)
\(\Leftrightarrow a^2+3a-4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=1\left(tm\right)\\a=-4\left(ktm\right)\end{matrix}\right.\)
\(\Rightarrow x^2+x+1=1\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\) \((tm)\)
Vậy \(S=\left\{0;-1\right\}\)
TL:
Giải thích các bước giải:
(2x-1) mũ 2 =9
(2x-1) mũ 2 =3 mũ 2
=> 2x-1= 3
2x =3+1
2x =4
x=4:2
x =2
vậy x = 2
\(\left(2x-1\right)^2=9\)
\(\Leftrightarrow\left(2x-1\right)^2-3^2=0\)
\(\Leftrightarrow\left(2x+2\right)\left(2x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+2=0\\2x-4=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)