Tìm x:
\(2^4.x-3.5x=5^2-2^4\)
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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
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
16x - 15x = 25 - 16
x = 9
2) 3².x + 2²x = 26.2² - 13
9x + 4x = 26.4 - 13
13x = 104 - 13
13x = 91
x = 91 : 13
x = 7
5) 6²x - 5²x = 11.2 - 11
36x - 25x = 22 - 11
11x = 11
x = 11 : 11
x = 1
3) 5²x - 2⁴x = 3⁴ - 16.3²
25x - 16x = 81 - 16.9
9x = -63
x = -63 : 9
x = -7
6) 7²x - 6²x = 13.2³ - 26
49x - 36x = 13.8 - 26
13x = 104 - 26
13x = 78
x = 78 : 13
x = 6
4) 7²x - 14x = 7².10 - 70
49x - 14x = 49.10 - 70
35x = 490 - 70
35x = 420
x = 420 : 35
x = 12
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
=>\(\left[{}\begin{matrix}3x+5=0\\x-5=0\end{matrix}\right.\)=>\(\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=5\end{matrix}\right.\)
16x - 3,5x = 9
=> x( 16 - 3,5 ) = 9
=> 12,5x = 9
=> x = 9 : 12,5
X = 0,72
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