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\(11x^2-15x+4=0\)
\(\Leftrightarrow11x^2-11x-4x+4=0\)
\(\Leftrightarrow11x\left(x-1\right)-4\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(11x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\11x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{4}{11}\end{matrix}\right.\)
\(S=\left\{1,\dfrac{4}{11}\right\}\)
Đặt C(x)=0
\(\Leftrightarrow11x^2-15x+4=0\)
\(\Leftrightarrow11x^2-11x-4x+4=0\)
\(\Leftrightarrow11x\left(x-1\right)-4\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(11x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\11x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\11x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{4}{11}\end{matrix}\right.\)
Vậy: Nghiệm của đa thức \(C\left(x\right)=11x^2-15x+4\) là 1 và \(\dfrac{4}{11}\)
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Ta có: x+y+1=0
nên x+y=-1
Ta có: \(N=x^2\left(x+y\right)-y^2\left(x+y\right)+x^2-y^2+2\left(x+y\right)+3\)
\(=\left(x+y\right)\left(x^2-y^2\right)+\left(x^2-y^2\right)+2\left(x+y\right)+3\)
\(=\left(x^2-y^2\right)\left(x+y+1\right)+2\left(x+y\right)+3\)
\(=\left(x^2-y^2\right)\cdot0+2\cdot\left(-1\right)+3\)
=-2+3=1
![](https://rs.olm.vn/images/avt/0.png?1311)
Đáp án:
P=\(\frac{2}{3}\)
Giải thích các bước giải:
x:y:z=5:4:3
⇒ x5x5 =y4y4 ⇒y= 4x54x5
⇒ x5x5 =z3z3 ⇒z= 3x53x5
Thay vào biểu thức ta được:
P= x+2y−3zx−2y+3zx+2y−3zx−2y+3z= x+2.4x5−33x5x−2.4x5+33x5x+2.4x5−33x5x−2.4x5+33x5 =4x56x54x56x5 =2323
Vậy P=\(\frac{2}{3}\)
# Chúc bạn học tốt!
Vì x,y,z tỉ lệ với các số 5,4,3 nên ta có : \(x:y:z=5:4:3\) hoặc \(\frac{x}{5}=\frac{y}{4}=\frac{z}{3}\)
Ta lại có : \(\frac{x}{5}=\frac{y}{4}=\frac{z}{3}=\frac{x}{5}=\frac{2y}{8}=\frac{3z}{9}\)
Đặt \(\frac{x}{5}=\frac{2y}{8}=\frac{3z}{9}=k\Rightarrow\hept{\begin{cases}x=5k\\2y=8k\\3z=9k\end{cases}}\)
\(P=\frac{x+2y-3z}{x-2y+3z}=\frac{5k+8k-9k}{5k-8k+9k}=\frac{4k}{6k}=\frac{4}{6}=\frac{2}{3}\)
Vậy \(P=\frac{2}{3}\)
13: \(\left(x-1\right)\left(x+1\right)=x^2+x-x-1=x^2-1\)
14: \(\left(x-5\right)\left(x+5\right)=x^2+5x-5x-25=x^2-25\)
15: \(\left(x-6\right)\left(6+x\right)\)
=(x-6)(x+6)
\(=x^2+6x-6x-36=x^2-36\)
16: \(\left(2x+1\right)\left(2x-1\right)=4x^2-2x+2x-1=4x^2-1\)
17: \(\left(x-2y\right)\left(x+2y\right)=x^2+2xy-2xy-4y^2=x^2-4y^2\)
18: \(\left(5x-3y\right)\cdot\left(3y+5x\right)\)
\(=\left(5x-3y\right)\left(5x+3y\right)\)
\(=25x^2+15xy-15xy-9y^2=25x^2-9y^2\)
19: \(\left(\dfrac{1}{x}-5\right)\left(\dfrac{1}{x}+5\right)=\left(\dfrac{1}{x}\right)^2+\dfrac{5}{x}-\dfrac{5}{x}-25=\dfrac{1}{x^2}-25\)
20: \(\left(x-\dfrac{3}{2}\right)\left(x+\dfrac{3}{2}\right)=x^2+\dfrac{3}{2}x-\dfrac{3}{2}x-\dfrac{9}{4}=x^2-\dfrac{9}{4}\)
21: \(\left(\dfrac{x}{3}-\dfrac{y}{4}\right)\left(\dfrac{x}{3}+\dfrac{y}{4}\right)=\left(\dfrac{x}{3}\right)^2+\dfrac{xy}{12}-\dfrac{xy}{12}-\left(\dfrac{y}{4}\right)^2\)
\(=\dfrac{x^2}{9}-\dfrac{y^2}{16}\)
22: \(\left(\dfrac{x}{y}-\dfrac{2}{3}\right)\left(\dfrac{x}{y}+\dfrac{2}{3}\right)=\left(\dfrac{x}{y}\right)^2+\dfrac{2}{3}\cdot\dfrac{x}{y}-\dfrac{2}{3}\cdot\dfrac{x}{y}-\left(\dfrac{2}{3}\right)^2\)
\(=\left(\dfrac{x}{y}\right)^2-\left(\dfrac{2}{3}\right)^2=\dfrac{x^2}{y^2}-\dfrac{4}{9}\)
23: \(\left(\dfrac{x}{2}+\dfrac{y}{3}\right)\left(\dfrac{y}{3}-\dfrac{x}{2}\right)=\left(\dfrac{y}{3}+\dfrac{x}{2}\right)\left(\dfrac{y}{3}-\dfrac{x}{2}\right)\)
\(=\left(\dfrac{y}{3}\right)^2-\dfrac{x}{2}\cdot\dfrac{y}{3}+\dfrac{x}{2}\cdot\dfrac{y}{3}-\left(\dfrac{x}{2}\right)^2\)
\(=\left(\dfrac{y}{3}\right)^2-\left(\dfrac{x}{2}\right)^2=\dfrac{y^2}{9}-\dfrac{x^2}{4}\)
24: \(\left(2x-\dfrac{2}{3}\right)\left(\dfrac{2}{3}+2x\right)=\left(2x-\dfrac{2}{3}\right)\left(2x+\dfrac{2}{3}\right)\)
\(=4x^2+\dfrac{4}{3}x-\dfrac{4}{3}x-\dfrac{4}{9}=4x^2-\dfrac{4}{9}\)