Giup tớ với mấy bạn ơi !!!!!!!!
Tìm x, biết :
a) \(3^{x-1}+5.3^{x-1}=162\)
b) \(3x+x^2=0\)
c) \(\left(x-1\right)\left(x-3\right)< 0\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
15 tháng 7 2021
a) Ta có: \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=15\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6\left(x^2+2x+1\right)=15\)
\(\Leftrightarrow-6x^2+12x+19+6x^2+12x+6=15\)
\(\Leftrightarrow24x+25=15\)
\(\Leftrightarrow24x=-10\)
hay \(x=-\dfrac{5}{12}\)
b) Ta có: \(2x^3-50x=0\)
\(\Leftrightarrow2x\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)
c) Ta có: \(5x^2-4\left(x^2-2x+1\right)-5=0\)
\(\Leftrightarrow5x^2-4x^2+8x-4-5=0\)
\(\Leftrightarrow x^2+8x-9=0\)
\(\Leftrightarrow\left(x+9\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=1\end{matrix}\right.\)
d) Ta có: \(x^3-x=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
e) Ta có: \(27x^3-27x^2+9x-1=1\)
\(\Leftrightarrow\left(3x\right)^3-3\cdot\left(3x\right)^2\cdot1+3\cdot3x\cdot1^2-1^3=1\)
\(\Leftrightarrow\left(3x-1\right)^3=1\)
\(\Leftrightarrow3x-1=1\)
\(\Leftrightarrow3x=2\)
hay \(x=\dfrac{2}{3}\)
26 tháng 12 2021
a) \(\Rightarrow\dfrac{1}{3}x\left(x-2\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
b) \(\Rightarrow\left(x+5\right)\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=-5\\x=1\end{matrix}\right.\)
c) \(\Rightarrow x\left(x^2-\dfrac{1}{9}\right)=0\Rightarrow x\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{1}{3}\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)
e) \(\Rightarrow\left(x+2\right)\left(x+2-x+2\right)=0\Rightarrow\left(x+2\right).4=0\Rightarrow x=-2\)
f) \(\Rightarrow x\left(2x-3\right)+2\left(2x-3\right)=0\Rightarrow\left(2x-3\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-2\end{matrix}\right.\)
g) \(\Rightarrow2\left(3x-2\right)^2-\left(3x-2\right)\left(3x+2\right)=0\Rightarrow\left(3x-2\right)\left(3x-6\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=2\end{matrix}\right.\)
h) \(\Rightarrow x\left(x+1\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=-2\end{matrix}\right.\)
i) \(\Rightarrow4x\left(x+1\right)+5\left(x+1\right)=0\Rightarrow\left(x+1\right)\left(4x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{5}{4}\end{matrix}\right.\)
20 tháng 7 2016
a.
\(\sqrt{2x+3}=1\)
\(2x+3=1\)
\(2x=1-3\)
\(2x=-2\)
\(x=-\frac{2}{2}\)
\(x=-1\)
b.
\(\left(3x-1\right)^2-25=0\)
\(\left(3x-1\right)^2=25\)
\(\left(3x-1\right)^2=\left(\pm5\right)^2\)
\(3x-1=\pm5\)
TH1:
\(3x-1=5\)
\(3x=5+1\)
\(3x=6\)
\(x=\frac{6}{3}\)
\(x=2\)
TH2:
\(3x-1=-5\)
\(3x=-5+1\)
\(3x=-4\)
\(x=-\frac{4}{3}\)
Vậy \(x=2\) hoặc \(x=-\frac{4}{3}\)
c.
\(\left(2x+4\right)\left(x^2+1\right)\left(x-2\right)=0\)
TH1:
\(2x+4=0\)
\(2x=-4\)
\(x=-\frac{4}{2}\)
\(x=-2\)
TH2:
\(x^2+1=0\)
\(x^2=-1\)
mà \(x^2\ge0\) với mọi x
=> loại
TH3:
\(x-2=0\)
\(x=2\)
Vậy \(x=2\) hoặc \(x=-2\)
20 tháng 7 2016
\(a.\)\(=>2x+3=1\)\(=>2x=-2\)\(=>x=-1\)
\(b.\)\(=>\left(3x-1\right)^2=25\)\(=>\left(3x-1\right)^2=5^2=>3x-1=5=>3x=6=>x=2\)
\(c.\)\(=>2x+4=0\)hoac \(x^2+1=0\)hoac \(x-2=0\)
=> * 2x=4 => x= 2
* x^2=-1=> x=-1
* x = 2
\(=>x\in\left(2;-1\right)\)
25 tháng 7 2021
a/ \(x=\dfrac{-5}{12}\)
b/ \(x\approx-1,9526\)
c/ \(x=\dfrac{21-i\sqrt{199}}{10}\)
d/ \(x=\dfrac{-20}{13}\)
25 tháng 7 2020
\(\frac{1}{3}.3^n+5.3^{n-1}=162\)
<=> \(3^{n-1}+5.3^{n-1}=162\)
<=> \(3^{n-1}\left(1+5\right)=162\)
<=> \(3^{n-1}.6=162\)
<=> \(3^{n-1}=162:6\)
<=> \(3^{n-1}=27\)
<=> \(3^{n-1}=3^3\)
<=> n - 1 = 3
<=> n = 3 + 1 = 4
Y
25 tháng 7 2020
Câu 1
a) Từ gt=>\(\hept{\begin{cases}x-5=1-3x\\x-5=3x-1\end{cases}}\)
<=>\(\hept{\begin{cases}4x=6\\2x=-4\end{cases}}\)
<=>\(\hept{\begin{cases}x=\frac{3}{2}\\x=-2\end{cases}}\)
b) Ta có: \(\hept{\begin{cases}\left(3x-1\right)^{100}\ge0,\forall x\in R\\\left(2y+1\right)^{200}\ge0,\forall x\in R\end{cases}}\)
Kết hợp với đề bài => \(\hept{\begin{cases}3x-1=0\\2y+1=0\end{cases}}\)
=>\(\hept{\begin{cases}x=\frac{1}{3}\\y=-\frac{1}{2}\end{cases}}\)
Bài 2
\(\frac{1}{3}.3^n+5.3^{n-1}=162\)
<=>\(3^{n-1}+5.3^{n-1}=162\)
<=>\(6.3^{n-1}=162\)
<=>\(3^{n-1}=27=3^3\)
<=>\(n-1=3\)
<=>\(n=4\)
26 tháng 1 2023
b: =>(2x-1)(2x-1+4-2x)=0
=>3(2x-1)=0
=>2x-1=0
=>x=1/2
c: =>(x+1)(x^2-x+1)-x(x+1)=0
=>(x+1)(x-1)^2=0
=>x=1 hoặc x=-1
e: =>(2x-1)(2x+1)=0
=>x=1/2 hoặc x=-1/2
h: =>x[(x^2-5)^2-4]=0
=>x(x^2-7)(x^2-3)=0
=>\(x\in\left\{0;\pm\sqrt{7};\pm\sqrt{3}\right\}\)
k: =>(x-1)(5x+3-3x+8)=0
=>(x-1)(2x+11)=0
=>x=1 hoặc x=-11/2
l: =>x^2(x+1)+(x+1)=0
=>(x+1)(x^2+1)=0
=>x+1=0
=>x=-1
E
15 tháng 9 2021
a)\(3x^2-4x=0<=>x(3x-4)=0\)
TH1: x=0
TH2 3x-4=0 <=>x=4/3
KL:.....
b) (x+3)(x−1)+2x(x+3)=0.
<=> (x+3)(x-1+2x)=0
TH1: x+3=0 <=> x=-3
TH2 x-1=0 <=> x=1
KL:.....
c) \(9x^2+6x+1=0. <=>(3x+1)^2=0<=>3x+1=0<=>x=-1/3 \)
KL:......
d) \(x^2−4x=4.<=>(x-2)^2=0<=>x-2=0<=>x=2\)
KL:....
15 tháng 9 2021
a) \(3x^2-4x=0\)
\(\Leftrightarrow x\left(3x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{3}\end{matrix}\right.\)
b) \(\left(x+3\right)\left(x-1\right)+2x\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{1}{3}\end{matrix}\right.\)
c) \(9x^2+6x+1=0\)
\(\Leftrightarrow\left(3x+1\right)^2=0\)
\(\Leftrightarrow3x+1=0\Leftrightarrow x=-\dfrac{1}{3}\)
d) \(x^2-4x=4\)
\(\Leftrightarrow\left(x-2\right)^2=8\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=2\sqrt{2}\\x-2=-2\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\sqrt{2}+2\\x=-2\sqrt{2}+2\end{matrix}\right.\)
a) 3x-1(1+5)=162
3x-1.6=162
3x-1=162:6=27=33
=>x-1=3
x=4
b) x(x+3)=0
=>x=0 hoặc x+3=0
=>x=0 hoặc x=-3
c) Vì tích nhỏ hơn 0 nên có 1 thừa số dương và 1 thừa số âm
Có x-1>x-3
=>x-1>0 và x-3<0
=>x>1 và x<3
Vậy x=2
a) 3x-1 + 5. 3x-1 = 162
1. 3x-1 + 5. 3x-1 = 162
( 1 + 5 ) . 3x-1 = 162
6. 3x-1 = 162
3x-1 = 162 : 6
3x-1 = 27
3x-1 = 33
x - 1 =3
x = 3 + 1
x = 4