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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 \(P=\frac{2}{3}\)
\(a,1\dfrac{1}{3}-\dfrac{3}{4}.\left(-\dfrac{8}{5}\right)+\left(-\dfrac{1}{5}\right)\)
\(=\dfrac{4}{3}-\left(-\dfrac{6}{5}\right)-\dfrac{1}{5}\)
\(=\dfrac{4}{3}+\dfrac{6}{5}-\dfrac{1}{5}=\dfrac{4}{3}+\left(\dfrac{6}{5}-\dfrac{1}{5}\right)\)
\(=\dfrac{4}{3}+1=\dfrac{7}{3}\)
\(b,\dfrac{5}{9}:2\dfrac{2}{5}-4\dfrac{5}{9}:2\dfrac{2}{5}=\dfrac{5}{9}:\dfrac{12}{5}-\dfrac{41}{9}:\dfrac{12}{5}\)
\(=\dfrac{5}{9}.\dfrac{5}{12}-\dfrac{41}{9}.\dfrac{5}{12}=\dfrac{5}{12}\left(\dfrac{5}{9}-\dfrac{41}{9}\right)\)\(=\dfrac{5}{9}.\left(-4\right)=-\dfrac{20}{9}\)
\(c,\left(-\dfrac{3}{4}+\dfrac{2}{3}\right):\dfrac{5}{11}+\left(-\dfrac{1}{4}+\dfrac{1}{3}\right):\dfrac{5}{11}\)
=\(\left(-\dfrac{3}{4}+\dfrac{2}{3}\right).\dfrac{11}{5}+\left(-\dfrac{1}{4}+\dfrac{1}{3}\right).\dfrac{11}{5}\)
=\(\dfrac{11}{5}\left(-\dfrac{3}{4}+\dfrac{2}{3}-\dfrac{1}{4}+\dfrac{1}{3}\right)\)
\(=\dfrac{11}{5}\left(-1+1\right)=\dfrac{11}{5}.0=0\)
d) \(\dfrac{1}{7}.\dfrac{3}{8}+\dfrac{1}{7}.\dfrac{5}{8}+\dfrac{\left(-1\right)^{2019}}{7}\)
=\(\dfrac{1}{7}\left(\dfrac{3}{8}+\dfrac{5}{8}\right)+\dfrac{-1}{7}=\dfrac{1}{7}.1-\dfrac{1}{7}\\\)
=1/7-1/7=0
\(e,-\dfrac{1}{8}.\dfrac{5}{6}-\dfrac{1}{8}.\dfrac{1}{6}+\dfrac{\left(-2020\right)^0}{8}\)
=\(-\dfrac{1}{8}\left(\dfrac{5}{6}+\dfrac{1}{6}\right)+\dfrac{1}{8}\)
\(=-\dfrac{1}{8}+\dfrac{1}{8}=0\)