Tìm x: 25%x+x=5/2
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a, => x^2+5 = 0
=> x^2=-5 ( vô lí vì x^2 >= 0)
=> ko tồn tại x tm bài toán
b, Vì x^2-5 > x^2-25
Mà (x^2-5): (x^2-25) < 0
=> x^2-5 >0 và x^2-25 <0
=> 5 < x^2 < 25
=> \(x>\sqrt{5}\)hoặc \(x< -\sqrt{5}\) và -5 < x < 5
=> -5 < x < -\(\sqrt{5}\)hoặc \(\sqrt{5}\)< x < 5
k mk nha
a) \(A\left(x\right)=x^2-10x+25\)
\(\Rightarrow A\left(x\right)=\left(x-5\right)^2\)
\(\Rightarrow\left\{{}\begin{matrix}A\left(0\right)=\left(0-5\right)^2=25\\A\left(-1\right)=\left(-1-5\right)^2=36\end{matrix}\right.\)
b) \(A\left(x\right)+B\left(x\right)=6x^2-5x+25\)
\(\Rightarrow B\left(x\right)=6x^2-5x+25-A\left(x\right)\)
\(\Rightarrow B\left(x\right)=6x^2-5x+25-\left(x^2-10x+25\right)\)
\(\Rightarrow B\left(x\right)=6x^2-5x+25-x^2+10x-25\)
\(\Rightarrow B\left(x\right)=5x^2+5x\)
\(\Rightarrow B\left(x\right)=5x\left(x+1\right)\)
c) \(A\left(x\right)=\left(x-5\right)C\left(x\right)\)
\(\Rightarrow C\left(x\right)=\dfrac{\left(x-5\right)^2}{x-5}=x-5\left(x\ne5\right)\)
d) Nghiệm của B(x)
\(\Leftrightarrow B=0\)
\(\Leftrightarrow5x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\) là nghiệm của B(x)
Bạn nên gõ đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người hiểu đề và hỗ trợ bạn tốt hơn nhé.
a. `4x^2-20x+25=0`
`<=>(2x)^2-2.2x.5 +5^2=0`
`<=>(2x-5)^2=0`
`<=>2x-5=0`
`<=>x=5/2`
b. `(x-5)(x+5)-(x-3)^2=2(x-7)`
`<=>x^2-25-x^2+6x-9=2x-14`
`<=>6x-34=2x-14`
`<=>4x=20`
`<=>x=5`
\(a,4x^2-20x+25=0\Leftrightarrow\left(2x\right)^2-2.2x.5+5^2=0\)
\(\Leftrightarrow\left(2x-5\right)^2=0\Leftrightarrow x=\dfrac{5}{2}\)
b, \(\left(x-5\right)\left(x+5\right)-\left(x-3\right)^2=2\left(x-7\right)\)
\(\Leftrightarrow x^2-25-x^2+6x-9=2x-14\Leftrightarrow4x=20\Leftrightarrow x=5\)
\(P=\left(\frac{x}{x^2-25}-\frac{x-5}{x^2+5x}\right):\frac{10x-25}{x^2+5x}+\frac{x}{5-x}\)
\(=\left[\frac{x}{\left(x-5\right)\left(x+5\right)}-\frac{x-5}{x\left(x+5\right)}\right]:\frac{10x-25}{x^2+5x}+\frac{x}{5-x}\)
\(=\left[\frac{x^2}{x\left(x-5\right)\left(x+5\right)}-\frac{\left(x-5\right)^2}{x\left(x-5\right)\left(x+5\right)}\right]:\frac{10x-25}{x^2+5x}+\frac{x}{5-x}\)
\(=\frac{x^2-\left(x^2-10x+25\right)}{x\left(x-5\right)\left(x+5\right)}:\frac{10x-25}{x\left(x+5\right)}+\frac{x}{5-x}\)
\(=\frac{10x-25}{x\left(x-5\right)\left(x+5\right)}.\frac{x\left(x+5\right)}{10x-25}+\frac{x}{5-x}\)
\(=\frac{1}{x-5}-\frac{x}{x-5}\)
\(=\frac{1-x}{x-5}=-\frac{x-1}{x-5}=-\frac{x-5+4}{x-5}=-1-\frac{4}{x-5}\)
Để P nguyên <=> x - 5 thuộc Ư(4) = {1;-1;2;-2;4;-4}
Ta có bảng:
x - 5 | 1 | -1 | 2 | -2 | 4 | -4 |
x | 6 | 4 | 7 | 3 | 9 | 1 |
Vậy....
a, (x2+5).(x2-25)=0
trường hợp 1(+):
x2+5=0
x2=0-5
x2=-5
không có giá trị nguyên nào
trường hợp 2
x2-25=0
x2=0+25
x2=25
x=5
vậy x =5
câu b làm tương tự nha. thông cảm
\(a,(x-2)^2-25=0\\\Leftrightarrow (x-2)^2=25\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)
\(---\)
\(b,4x(x-2)+x-2=0\\\Leftrightarrow4x(x-2)+(x-2)=0\\\Leftrightarrow(x-2)(4x+1)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\4x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{1}{4}\end{matrix}\right.\)
\(---\)
\(c,4x(x-2)-x(3+4x)(?)\)
\(d,(2x-5)^2-3x(5-2x)=0\\\Leftrightarrow(2x-5)^2+3x(2x-5)=0\\\Leftrightarrow(2x-5)(2x-5+3x)=0\\\Leftrightarrow(2x-5)(5x-5)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\5x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=1\end{matrix}\right.\)
\(---\)
\(e,x^2-25-(x+5)=0(sửa.đề)\\\Leftrightarrow(x^2-5^2)-(x+5)=0\\\Leftrightarrow (x-5)(x+5)-(x+5)=0\\\Leftrightarrow(x+5)(x-5-1)=0\\\Leftrightarrow(x+5)(x-6)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=6\end{matrix}\right.\)
\(---\)
\(f,5x(x-3)-x+3=0\\\Leftrightarrow5x(x-3)-(x-3)=0\\\Leftrightarrow(x-3)(5x-1)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\5x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)
\(Toru\)
\(\left(x^2-25\right)^2-\left(x+5\right)^2=0\)
\(\Leftrightarrow\left[x^2-5^2\right]^2-\left(x+5\right)^2=0\)
\(\Leftrightarrow\left[\left(x+5\right)\left(x-5\right)\right]^2-\left(x+5\right)^2=0\)
\(\Leftrightarrow\left(x+5\right)^2\left(x-5\right)^2-\left(x+5\right)^2=0\)
\(\Leftrightarrow\left(x+5\right)^2\left[\left(x-5\right)^2-1\right]=0\)
\(\Leftrightarrow\left(x+5\right)^2\left[\left(x-5\right)+1\right]\left[\left(x-5\right)-1\right]=0\)
\(\Leftrightarrow\left(x+5\right)^2\left(x-4\right)\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x+5\right)^2=0\\x-4=0\\x-6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x=4\\x=6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=4\\x=6\end{matrix}\right.\)
Vậy: \(S=\left\{-5;6;4\right\}\)
Ta có ( x2 - 25 )2 - ( x + 5 )2 = 0
Vì ( x2 - 25 )2 ≥ 0 ; ( x + 5 )2 ≥ 0
⇒ ( x2 - 25 )2 - ( x + 5 )2 ≥ 0
Dấu " = " xảy ra khi
\(\left[{}\begin{matrix}\left(x^2-25\right)^2=0\\\left(x+5\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\pm5\\x=-5\end{matrix}\right.\Rightarrow x=-5\)
Vậy x = 5
\(25\%x+x=\frac{5}{2}\)
\(\frac{1}{4}x+1x=\frac{5}{2}\)
\(\left(\frac{1}{4}+1\right)x=\frac{5}{2}\)
\(\frac{5}{4}x=\frac{5}{2}\)
\(x=\frac{5}{2}:\frac{5}{4}\)
\(x=2\)
Thank you bạn nhé