5x+4 -3.5x+3= 2.511. Tìm x.
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\(1,x^3-3x^2=0\)
\(x^2\left(x-3\right)=0\)
\(\orbr{\begin{cases}x^2=0\\x-3=0\end{cases}\orbr{\begin{cases}x=0\left(TM\right)\\x=3\left(TM\right)\end{cases}}}\)
\(2,3x^3-48x=0\)
\(3x\left(x^2-16\right)=0\)
\(\orbr{\begin{cases}3x=0\\x^2-16=0\end{cases}\orbr{\begin{cases}x=0\left(TM\right)\\x^2=16\end{cases}\orbr{\begin{cases}x=0\left(TM\right)\\x=\pm4\left(TM\right)\end{cases}}}}\)
\(3,5x\left(x-1\right)=x-1\)
\(5x^2-5x=x-1\)
\(5x^2-6x+1=0\)
\(5x^2-5x-x+1=0\)
\(5x\left(x-1\right)-\left(x-1\right)=0\)
\(\left(5x-1\right)\left(x-1\right)=0\)
\(\orbr{\begin{cases}5x-1=0\\x-1=0\end{cases}\orbr{\begin{cases}x=\frac{1}{5}\left(TM\right)\\x=1\left(TM\right)\end{cases}}}\)
\(4,2\left(x+5\right)-x^2-5x=0\)
\(2x+10-x^2-5x=0\)
\(-x^2-3x+10=0\)
\(-x^2-5x+2x+10=0\)
\(-x\left(x+5\right)+2\left(x+5\right)=0\)
\(\left(x+5\right)\left(2-x\right)=0\)
\(\orbr{\begin{cases}x+5=0\\2-x=0\end{cases}\orbr{\begin{cases}x=-5\left(TM\right)\\x=2\left(TM\right)\end{cases}}}\)
\(5,2x\left(x-5\right)-x\left(3+2x\right)=26\)
\(2x^2-10x-3x-2x^2=26\)
\(-13x-26=0\)
\(-13\left(x+2\right)=0\)
\(x=-2\left(TM\right)\)
Trả lời:
1, \(x^3-3x^2=0\)
\(\Leftrightarrow x^2\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}}\)
Vậy x = 0; x = 3 là nghiệm của pt.
2, \(3x^3-48x=0\)
\(\Leftrightarrow3x\left(x^2-16\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x=0\\x^2-16=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm4\end{cases}}}\)
Vậy x = 0; x = 4; x = - 4 là nghiệm của pt.
3, \(5x\left(x-1\right)=x-1\)
\(\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\5x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{5}\end{cases}}}\)
Vậy x = 1; x = 1/5 là nghiệm của pt.
4, \(2\left(x+5\right)-x^2-5x=0\)
\(\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(2-x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+5=0\\2-x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=2\end{cases}}}\)
Vậy x = - 5; x = 2 là nghiệm của pt.
5, \(2x\left(x-5\right)-x\left(3+2x\right)=26\)
\(\Leftrightarrow2x^2-10x-3x-2x^2=26\)
\(\Leftrightarrow-13x=26\)
\(\Leftrightarrow x=-2\)
Vậy x = - 2 là nghiệm của pt.
16. 2 x + 4. 2 x = 5. 5 x + 3. 5 x
⇔ 20. 2 x = 8. 5 x ⇔ 2 / 5 x = 2 / 5 1 ⇔ x = 1
Bài 3:
1. \(\left(x-1\right)\left(x+2\right)+5x-5=0\)
\(\Rightarrow\left(x-1\right)\left(x+2\right)+5\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(x+2+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x+7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)
Vậy.......................
2. \(\left(3x+5\right)\left(x-3\right)-6x-10=0\)
\(\Rightarrow\left(3x+5\right)\left(x-3\right)-2\left(3x+5\right)=0\)
\(\Rightarrow\left(3x+5\right)\left(x-3-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}3x+5=0\\x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=5\end{matrix}\right.\)
Vậy........................
3. \(\left(x-2\right)\left(2x+3\right)-7x^2+14x=0\)
\(\Rightarrow\left(x-2\right)\left(2x+3\right)-7x\left(x-2\right)=0\)
\(\Rightarrow\left(x-2\right)\left(2x+3-7x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\-5x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{3}{5}\end{matrix}\right.\)
Vậy............................
4, 5 tương tự nhé bn!
bài 3
1 (x-1)(x+2)+5x-5=0
=>(x-1)(x+2)+(5x-5)=o
=>(x-1)(x+2)+5(x-1)=0
=>(x-1)(x+2+5)=0
=>(x-1)(x+7)=0
=>\(\left[{}\begin{matrix}x-1=0\\x+7=0\end{matrix}\right.\) =>\(\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)
vậy x=1 hoặc x=-7
2. (3x+5)(x-3)-6x-10=0
=>(3x+5)(x-3)-(6x+10)=0
=>(3x+5)(x-3)-2(3x+5)=0
=>(3x+5)(x-3-2)=0
=>(3x+5)(x-5)=0
=>\(\left[{}\begin{matrix}3x+5=0\\x-5=0\end{matrix}\right.\)=>\(\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=5\end{matrix}\right.\)
1) \(x\left(x+4\right)\left(x-4\right)-\left(x^2+1\right)\left(x^2-1\right)\)
\(=x\left(x^2-16\right)\)
\(=x^3-16x-\left(x^2+1\right)\left(x^2-1\right)\)
\(=x^3-16x-x^4+1\)
b) \(7x\left(4y-x\right)+4y\left(y-7x\right)-2\left(2y^2-3.5x\right)\)
\(=28xy-7x^2+4y\left(y-7x\right)-2\left(2y^2-3.5x\right)\)
\(=28xy-7x^2+4y^2-28xy-4y^2+7x\)
\(=-7x^2+7x\)
c) \(\left(3x-1\right)\left(2x-5\right)-4\left(2x^2-5x+2\right)\)
\(=6x^2-17x+5-4\left(2x^2-5x+2\right)\)
\(=6x^2-17x+5-8x^2+20x-8\)
\(=-2x^2+3x-3\)
a) x(x+4)(x-4)-(x2+1)(x2-1)
=>x(x2-42)-(x4-12)
=>x3-16x-x4+1
=>-x4-x3-15x
b) 7x(4y-x)+4y(y-7x)-2(2y2-3.5x)
=>28xy-7x2+4y2-28xy-4y2+30x
=>-7x2+30x
c) (3x+1)(2x-5)-4(2x2-5x+2)
=>6x2-15x+2x-5-8x2+20x-8
=>-2x2+7x-13
1) 24 . x - 3 . 5x = 52 - 24
16 . x - 3 . 5x = 25 - 16
16 . x - 3 . 5x = 9
Tự làm tiếp
2) 32 . x + 22 . x = 26 . 22 - 13
9 . x + 4 . x = 26 . 4 - 13
( 9 + 4 ) . x = 104 - 13
13 . x = 91
x = 91 : 13
x = 7
24.x - 3.5x = 52 - 24
=> 16.x - 15x = 25 - 16
=> x = 9
32.x + 22.x = 26.22 - 13
=> 9.x + 4.x = 26.4 - 13
=> 13.x = 91
=> x = 7
@Huỳnh Quang Sang bạn giải thích hộ mình tại sao lại ra được kết quả như vậy ko ạ, mình chưa hiểu rõ lắm, mong bạn giải đáp
\(5^{x+4}-3.5^{x+3}=2.5^{11}\)
\(\Leftrightarrow5^{x+3+1}-3.5^{x+3}=2.5^{11}\)
\(\Leftrightarrow5^{x+3}.5-3.5^{x+3}=2.5^{11}\)
\(\Leftrightarrow5^{x+3}.\left(5-3\right)=5^{11}.2\)
\(\Leftrightarrow5^{x+3}.2=5^{11}.2\)
\(\Leftrightarrow5^{x+3}=5^{11}\)
\(\Leftrightarrow x+3=11\)
\(\Leftrightarrow x=8\)
Vậy \(x=8\)