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)

một bể cá dạng hình hộp chữ nhật có chiều dài 1 2 m chiều rộng 0 5 m bên trong có một hòn non bộ  đặc có thể thích bằng 0,09m3 nếu đổ vào 150l nước thì hòn non bộ ngập hoàn toàn trong nước . hỏi chiều cao mực nước ở trong bẻ là bao nhiêu

a) \(7x\left(x+1\right)-3\left(x+1\right)=0\Rightarrow\left(x+1\right)\left(7x-3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+1=0\\7x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{3}{7}\end{matrix}\right.\)

b) 3(x + 8) - x² - 8x = 0

=> 3(x + 8) - (x² + 8x) = 0

=> 3(x + 8) - x(x + 8) = 0

=> (x + 8)(3 - x) = 0 => \(\left[{}\begin{matrix}x+8=0\\3-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-8\\x=3\end{matrix}\right.\)

c) \(x^2-10x=-25\Rightarrow x^2-10x+25=0\Rightarrow\left(x-5\right)^2=0\Rightarrow x=5\)

d) Giống câu c

⇒[x+1=07x+3=0⇒⎡⎣x=−1x=−37⇒[x+1=07x+3=0⇒[x=−1x=−37

b) 3(x + 8) - x² - 8x = 0

=> 3(x + 8) - (x² + 8x) = 0

=> 3(x + 8) - x(x + 8) = 0

=> (x + 8)(3 - x) = 0 => [x+8=03−x=0⇒[x=−8x=3[x+8=03−x=0⇒[x=−8x=3

c) ${}^{}-10x=-25\Rightarrow {}^{}-10x+$

a: =>5(x+2)=170

=>x+2=34

=>x=32

b: =>x+2=5 hoặc x+2=-5

=>x=-7 hoặc x=3

c: =>x-2=1 hoặc x-2=-1

=>x=1 hoặc x=3

Giúp em đi ạ:(

xem lại đề hộ mk nhé 

mk sửa đề thế này 

x+ ( x+1)+(x+2)+....+(x+24)+(x+25) = 25 

=> x+ x+1 + x + 2 +...+ x+ 24 + x + 25 = 25 

=> 25x + 325 = 25 

từ đây tính ra x

\(3\left(x-2\right)+4\left(x-1\right)=25\) 

\(\Leftrightarrow3x-6+4x-4=25\) 

\(\Leftrightarrow7x=35\) 

\(\Leftrightarrow x=5\)

\(\left(5x-3\right)\left(x-2\right)=\left(x-1\right)\left(x-2\right)\) 

\(\Leftrightarrow\left(5x-3\right)\left(x-2\right)-\left(x-1\right)\left(x-2\right)=0\) 

\(\Leftrightarrow\left(x-2\right)\left(5x-3-x+1\right)=0\) 

\(\Leftrightarrow\left(x-2\right)\left(4x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\4x+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-1}{2}\end{matrix}\right.\)

\(\left(x-2\right)^2=4\left(x-1\right)^2\)

\(\Leftrightarrow\left(x-2\right)^2-4\left(x-1\right)^2=0\) 

\(\Leftrightarrow\left[\left(x-2\right)-2\left(x-1\right)\right]\left[\left(x-2\right)+2\left(x-1\right)\right]=0\) 

\(\Leftrightarrow\left(x-2-2x+2\right)\left(x-2+2x-2\right)=0\) 

\(\Leftrightarrow\left(-x\right)\left(3x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-x=0\\3x-4=0\end{matrix}\right.\) 

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{3}\end{matrix}\right.\)

\(3\left(x-1\right)^2+\left(x+5\right)\left(2-3x\right)=-25\)

\(\Leftrightarrow3x^2-6x+3-3x^2-13x+10=-25\)

\(\Leftrightarrow-19x=-38\Leftrightarrow x=2\)

\(\Rightarrow3x^2-6x+3+2x-3x^2+10-15x=-25\\ \Rightarrow-19x=-38\\ \Rightarrow x=2\)

\(\left(x+1\right)^2=25\)

\(\Rightarrow\left(x+1\right)^2=\left(\pm5\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x+1=5\\x+1=-5\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=5-1\\x=-5-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=4\\x=-6\end{matrix}\right.\)

\(\left(x+1\right)\cdot2=25\)

\(2x+2=25\)

\(2x=25-2=23\)

\(x=23:2=11,5\)

\(\left(x+1\right)2=25\)

\(x+1=25:2\)

\(x+1=12,5\)

\(x=12,5-1\\ x=11,5\)

Vậy:\(x=11,5\)

a/

\(x^2=25\Leftrightarrow x=\pm5\)

b/

\(x^2-1=15\\\Leftrightarrow x^2=16\Leftrightarrow x=\pm4\)

c/

\(19-2x^2=1\Leftrightarrow2x^2=18\Leftrightarrow x^2=9\Leftrightarrow x=\pm3\)

`@` `\text {Ans}`

`\downarrow`

`a,`

`x^2 = 25`

`=> x^2 = (+-5)^2`

`=> x = +-5`

Vậy, `x \in {5; -5}`

`b,`

`x^2 - 1 = 15`

`=> x^2 = 15+1`

`=> x^2 = 16`

`=> x^2 = (+-4)^2`

`=> x = +-4`

Vậy, `x \in {4; -4}`

`c,`

`19 - 2x^2 = 1`

`=> 2x^2 = 19 - 1`

`=> 2x^2 = 18`

`=> x^2 = 18 \div 2`

`=> x^2 = 9`

`=> x^2 = (+-3)^2`

`=> x = +-3`

Vậy, `x \in {3; -3}.`

1, \(\left(x-1\right)\left(x+2\right)-\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(x-1\right)\left[x+2-\left(x-1\right)\right]=0\)

\(\Leftrightarrow3\left(x-1\right)=0\Leftrightarrow x=1\)

2, \(\left(x-2\right)^2-3\left(x-2\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left[x-2-3\left(x+1\right)\right]=0\)

\(\Leftrightarrow\left(x-2\right)\left(-2x-5\right)=0\Leftrightarrow x=-\dfrac{5}{2};x=2\)

3, \(\left(5-2x\right)\left(2x+7\right)=4x^2-25=\left(2x-5\right)\left(2x+5\right)\)

\(\Leftrightarrow\left(5-2x\right)\left(2x+7\right)+\left(5-2x\right)\left(2x+5\right)=0\)

\(\Leftrightarrow\left(5-2x\right)\left(2x+7+2x+5\right)=0\Leftrightarrow\left(4x+12\right)\left(5-2x\right)=0\Leftrightarrow x=-3;x=\dfrac{5}{2}\)

1) Ta có: \(\left(x-1\right)\left(x+2\right)-\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2-x+1\right)=0\)

\(\Leftrightarrow x-1=0\)

hay x=1

2) Ta có: \(\left(x-2\right)^2-3\left(x-2\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-2-3x-3\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(-2x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-5}{2}\end{matrix}\right.\)

a) \(\left(x-2\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)

b) \(\left(x^2+5\right)\left(x^2-25\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+5=0\\x^2-25=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2=-5\\x^2=25\end{matrix}\right.\) \(\Leftrightarrow x^2=25\) \(\Leftrightarrow x=\pm5\)