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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 \(\dfrac{4}{11}\)
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
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 \(P=\frac{2}{3}\)
\(a.\left(4x+1\right)\left(-2x+\dfrac{1}{3}\right)=0\\ TH1:4x+1=0\\ =>4x=-1\\ =>x=-\dfrac{1}{4}\\ TH2:-2x+\dfrac{1}{3}=0\\ =>2x=\dfrac{1}{3}\\ =>x=\dfrac{1}{3}:2=\dfrac{1}{6}\\ b.\left(x-\dfrac{5}{2}\right)^3=\dfrac{-1}{8}\\ =>\left(x-\dfrac{5}{2}\right)^3=\left(-\dfrac{1}{2}\right)^2\\ =>x-\dfrac{5}{2}=-\dfrac{1}{2}\\ =>x=-\dfrac{1}{2}+\dfrac{5}{2}\\ =>x=\dfrac{4}{2}=2\\ c.\left(\dfrac{2}{5}-3x\right)^2-\dfrac{1}{5}=\dfrac{4}{25}\\ =>\left(\dfrac{2}{5}-3x\right)^2=\dfrac{4}{25}+\dfrac{1}{5}=\dfrac{9}{25}=\left(\dfrac{3}{5}\right)^2\\TH1:\dfrac{2}{5}-3x=\dfrac{3}{5}\\ =>3x=\dfrac{2}{5}-\dfrac{3}{5}=-\dfrac{1}{5}\\ =>x=\dfrac{-1}{5}:3=-\dfrac{1}{15}\\ TH2:\dfrac{2}{5}-3x=-\dfrac{3}{5}=>3x=\dfrac{2}{5}+\dfrac{3}{5}=1\\ =>x=1:3=\dfrac{1}{3}\)
\(d.\left(\dfrac{2}{3}\right)^{x+2}+\left(\dfrac{2}{3}\right)^{x+1}=\dfrac{20}{27}\\ =>\left(\dfrac{2}{3}\right)^{x+1}\cdot\left(\dfrac{2}{3}+1\right)=\dfrac{20}{27}\\ =>\left(\dfrac{2}{3}\right)^{x+1}\cdot\dfrac{5}{3}=\dfrac{20}{27}\\ =>\left(\dfrac{2}{3}\right)^{x+1}=\dfrac{20}{27}:\dfrac{5}{3}=\dfrac{4}{9}=\left(\dfrac{2}{3}\right)^2\\ =>x+1=2\\ =>x=2-1\\ =>x=1\)