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

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

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

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

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

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

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

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:(

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

\(\Leftrightarrow3x-6+4x-4=25\) 

\(\Leftrightarrow7x=35\) 

\(\Leftrightarrow x=5\)

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

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

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

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

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

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

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

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

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

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

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

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

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

(\(x\) + 1)² = \(\dfrac{4}{25}\)

(\(x+1\))² = (\(\dfrac{2}{5}\))²

\(\left[{}\begin{matrix}x+1=-\dfrac{2}{5}\\x+1=\dfrac{2}{5}\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=\dfrac{2}{5}-1\\x=-\dfrac{2}{5}-1\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=-\dfrac{3}{5}\\x=-\dfrac{7}{5}\end{matrix}\right.\)

Vậy \(x\in\){  \(-\dfrac{7}{5}\) ; - \(\dfrac{3}{5}\)} 

`@` `\text {Ans}`

`\downarrow`

`(x+1)^2 = 4/25`

`=> (x+1)^2 = (+-2/5)^2`

`=>`\(\left[{}\begin{matrix}x+1=\dfrac{2}{5}\\x+1=-\dfrac{2}{5}\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=\dfrac{2}{5}-1\\x=-\dfrac{2}{5}-1\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=-\dfrac{3}{5}\\x=-\dfrac{7}{5}\end{matrix}\right.\)

Vậy, `x \in {-3/5; -7/5}.`

x + 25%x = \(\frac{1}{2}\)

x.1+\(\frac{1}{4}\)x = \(\frac{1}{2}\)

x.(\(1+\frac{1}{4}\)) = \(\frac{1}{2}\)

x . \(\frac{5}{4}\)= \(\frac{1}{2}\)

x       = \(\frac{1}{2}\div\frac{5}{4}\) 

x       = \(\frac{2}{5}\)

Ta có : \(X+25\%.X=\frac{1}{2}\Leftrightarrow X+\frac{1}{4}.X=\frac{1}{2}\Leftrightarrow X.\left(1+\frac{1}{4}\right)=\frac{1}{2}\)\(\Leftrightarrow X.\frac{5}{4}=\frac{1}{2}\Leftrightarrow X=\frac{1}{2}:\frac{5}{4}\Leftrightarrow X=\frac{2}{5}\)

bn qên chưa rút gọn rồi đó ngô mih hoàng à!

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}.`