tìm (x mũ 2 -1)(x=3)=0
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Bài 5 :
a, \(2x\left(x-3\right)+x-3=0\Leftrightarrow\left(2x+1\right)\left(x-3\right)=0\Leftrightarrow x=-\frac{1}{2};x=3\)
b, \(x\left(x+1\right)-x-1=0\Leftrightarrow\left(x-1\right)\left(x+1\right)=0\Leftrightarrow x=\pm1\)
c, sửa đề \(x^3-3x^2+x-3=0\Leftrightarrow x^2\left(x-3\right)+x-3=0\)
\(\Leftrightarrow\left(x^2+1>0\right)\left(x-3\right)=0\Leftrightarrow x=3\)
d, \(3x^2\left(2x-1\right)+1-4x^2=0\Leftrightarrow3x^2\left(2x-1\right)+\left(1-2x\right)\left(1+2x\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(3x^2-2x-1\right)=0\Leftrightarrow\left(2x-1\right)\left(3x+1\right)\left(x-1\right)=0\Leftrightarrow x=1;x=-\frac{1}{3};x=\frac{1}{2}\)
e, \(x^3+2x-x^2-2=0\Leftrightarrow x\left(x^2+2\right)-\left(x^2+2\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+2>0\right)=0\Leftrightarrow x=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1, \(x^3+4x^2+4x=0\Leftrightarrow x\left(x^2+4x+4\right)=0\)
\(\Leftrightarrow x\left(x+2\right)^2=0\Leftrightarrow x=-2;x=0\)
2, \(\left(x+3\right)^2-4=0\Leftrightarrow\left(x+3-2\right)\left(x+3+2\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+5\right)=0\Leftrightarrow x=-5;x=1\)
3, \(x^4-9x^2=0\Leftrightarrow x^2\left(x^2-9\right)=0\)
\(\Leftrightarrow x^2\left(x-3\right)\left(x+3\right)=0\Leftrightarrow x=0;\pm3\)
4, \(x^2-6x+9=81\Leftrightarrow\left(x-3\right)^2=9^2\)
\(\Leftrightarrow\left(x-3-9\right)\left(x-3+9\right)=0\Leftrightarrow\left(x-12\right)\left(x+6\right)=0\Leftrightarrow x=-6;x=12\)
5, em xem lại đề nhé
à lag tý @@
5, \(x^3+6x^2+9x-4x=0\Leftrightarrow x^3+6x^2+5x=0\)
\(\Leftrightarrow x\left(x^2+6x+5\right)=0\Leftrightarrow x\left(x^2+x+5x+5\right)=0\)
\(\Leftrightarrow x\left(x+1\right)\left(x+5\right)=0\Leftrightarrow x=-5;x=-1;x=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a. x mũ 2 - 2x + 1 = 25
= x^2 + 2.x.1 + 1^2
= ( x + 1 ) ^2
ko bt có đúng ko nữa, mấy câu kia tui ko bt lm
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\text{a) ( x + 1 ) + ( x + 3 ) + ..... + ( x + 99 ) = 0}\)
\(\Rightarrow\left(x+x+x+.....+x\right)+\left(1+3+5+....+99\right)=0\)
\(\text{Ta có :}\)
\(1+3+5+...+99=\frac{\left(99-1\right):2+1.\left(99+1\right)}{2}=2500\)
\(\Rightarrow50x+2500=0\)
\(\Rightarrow50x=-2500\)
\(\Rightarrow x=-50\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1, \(3x\left(x-7\right)+2x-14=0\)
\(\Rightarrow3x\left(x-7\right)+2\left(x-7\right)=0\)
\(\Rightarrow\left(x-7\right)\left(3x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=7\\x=\frac{-2}{3}\end{cases}}\)
2, \(x^3+3x^2-\left(x+3\right)=0\)
\(\Rightarrow x^2\left(x+3\right)-\left(x+3\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x^2-1\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=-3\\x=\pm1\end{cases}}\)
3, \(15x-5+6x^2-2x=0\)
\(\Rightarrow\left(15x-5\right)+\left(6x^2-2x\right)=0\)
\(\Rightarrow5\left(3x-1\right)+2x\left(3x-1\right)=0\)
\(\Rightarrow\left(3x-1\right)\left(5+2x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-1=0\\5+2x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=\frac{-5}{2}\end{cases}}\)
4, \(5x-2-25x^2+10x=0\)
\(\Rightarrow\left(5x-25x^2\right)-\left(2-10x\right)=0\)
\(\Rightarrow5x\left(1-5x\right)-2\left(1-5x\right)=0\)
\(\Rightarrow\left(1-5x\right)\left(5x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}1-5x=0\\5x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{2}{5}\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(\left(4-x\right)^2-16=0\)
\(\Leftrightarrow\left(4-x\right)^2=16\)
\(\Leftrightarrow\orbr{\begin{cases}4-x=4\\4-x=-4\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=8\end{cases}}}\)
b) \(25-\left(3-x\right)^2=0\)
\(\Leftrightarrow\left(3-x\right)^2=25\)
\(\Leftrightarrow\orbr{\begin{cases}3-x=5\\3-x=-5\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=8\end{cases}}}\)
c)\(3x^2-6x+3-27=0\)
\(\Leftrightarrow3x^2-6x-24=0\)
\(\Leftrightarrow\left(3x+6\right)\left(x-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x+6=0\\x-4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=4\end{cases}}}\)
#H
1.(4-x)2-16 =0
<=> 16 -8x+x2 -16 =0
<=> -x(8-x) =0
<=> TH1: x=0
. TH2: 8-x=0
. => x= -8
2. 25 - (3-x)2 = 0
<=> 25 - (9-6x+x2) = 0
<=> 25 - 9+6x-x2 = 0
<=> -x2+6x+16 = 0
<=> -(x-8)(x+2) = 0 (bước này bạn nhập phương trình trên mtinh là nó sẽ ra nghiệm nhe)
<=> TH1:x-8=0
. x= 8
. TH2:x+2=0
. x=-2
3.(bạn tự làm nhé, giải bth thui)
![](https://rs.olm.vn/images/avt/0.png?1311)
(x+3)^2 + (x-15)^2 = 0
co (x + 3^2) > 0 va (x-15)^2 > 0
=> (x+3)^2 = 0 va (x - 15)^2 = 0
=> x + 3 = 0 va x - 15 = 0
=> x = -3 va x = 15
vay x thuoc tap hop rong :v
Bạn phuong uyen không phải và đâu mà là hoặc đấy chỉ cần 1 trong hai cái =0
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Trả lời:
\(1,\left(4x-x\right)^2-16=0\)
\(\Leftrightarrow\left(3x\right)^2-16=0\)
\(\Leftrightarrow\left(3x-4\right)\left(3x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-4=0\\3x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{4}{3}\\x=-\frac{4}{3}\end{cases}}}\)
Vậy x = 4/3; x = - 4/3 là nghiệm của pt.
\(2,25-\left(3-x\right)^2=0\)
\(\Leftrightarrow\left(5-3+x\right)\left(5+3-x\right)=0\)
\(\Leftrightarrow\left(2+x\right)\left(8-x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2+x=0\\8-x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=8\end{cases}}}\)
Vậy x = - 2; x = 8 là nghiệm của pt.
\(3,3x^2-6x+3-27=0\)
\(\Leftrightarrow3x^2-6x-24=0\)
\(\Leftrightarrow3\left(x^2-2x-8\right)=0\)
\(\Leftrightarrow x^2-2x-8=0\)
\(\Leftrightarrow x^2-4x+2x-8=0\)
\(\Leftrightarrow x\left(x-4\right)+2\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x+2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=-2\end{cases}}\)
Vậy x = 4; x = - 2 là nghiệm của pt.
Ta có :
sao chỗ này bn ghi x= 3 thế là - hay + để mk còn giải