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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 \(P=\frac{2}{3}\)
Bài 5:
\(a)\dfrac{3}{5}\cdot\dfrac{6}{7}+\dfrac{3}{7}:\dfrac{5}{3}-\dfrac{2}{7}:1\dfrac{2}{3}\\ =\dfrac{3}{5}\cdot\dfrac{6}{7}+\dfrac{3}{7}\cdot\dfrac{3}{5}-\dfrac{2}{7}:\dfrac{5}{3}\\ =\dfrac{3}{5}\cdot\dfrac{6}{7}+\dfrac{3}{7}\cdot\dfrac{3}{5}-\dfrac{2}{7}\cdot\dfrac{3}{5}\\ =\dfrac{3}{5}\cdot\left(\dfrac{6}{7}+\dfrac{3}{7}-\dfrac{2}{7}\right)\\ =\dfrac{3}{5}\cdot\dfrac{7}{7}\\=\dfrac{3}{5}\)
\(b)\dfrac{4}{9}:\left(\dfrac{1}{15}-\dfrac{2}{3}\right)+\dfrac{4}{9}:\left(\dfrac{1}{11}-\dfrac{5}{22}\right)\\ =\dfrac{4}{9}:\left(\dfrac{1}{15}-\dfrac{10}{15}\right)+\dfrac{4}{9}:\left(\dfrac{2}{22}-\dfrac{5}{22}\right)\\ =\dfrac{4}{9}:\dfrac{-9}{15}+\dfrac{4}{9}:\dfrac{-3}{22}\\ =\dfrac{4}{9}\cdot\dfrac{-5}{3}+\dfrac{4}{9}\cdot\dfrac{-22}{3}\\ =\dfrac{4}{9}\cdot\left(\dfrac{-5}{3}+\dfrac{-22}{3}\right)\\ =\dfrac{4}{9}\cdot\dfrac{-27}{3}\\ =\dfrac{4}{9}\cdot\left(-9\right)\\ =-4\)
Bài 4:
a: \(-1\dfrac{2}{5}\cdot x=\dfrac{2}{3}-\dfrac{4}{5}\)
=>\(x\cdot\dfrac{-7}{5}=\dfrac{10}{15}-\dfrac{12}{15}=-\dfrac{2}{15}\)
=>\(x=\dfrac{-2}{15}:\dfrac{-7}{5}=\dfrac{2}{15}\cdot\dfrac{5}{7}=\dfrac{2}{21}\)
b: \(\dfrac{2}{3}-4x=\dfrac{1}{2}-\dfrac{2}{5}\)
=>\(\dfrac{2}{3}-4x=\dfrac{5}{10}-\dfrac{4}{10}=\dfrac{1}{10}\)
=>\(4x=\dfrac{2}{3}-\dfrac{1}{10}=\dfrac{20}{30}-\dfrac{3}{30}=\dfrac{17}{30}\)
=>\(x=\dfrac{17}{30}:4=\dfrac{17}{120}\)
Bài 5:
a: \(\dfrac{3}{5}\cdot\dfrac{6}{7}+\dfrac{3}{7}:\dfrac{5}{3}-\dfrac{2}{7}:1\dfrac{2}{3}\)
\(=\dfrac{3}{5}\cdot\dfrac{6}{7}+\dfrac{3}{7}\cdot\dfrac{3}{5}-\dfrac{2}{7}:\dfrac{7}{3}\)
\(=\dfrac{18+9}{35}-\dfrac{2}{7}\cdot\dfrac{3}{7}=\dfrac{27}{35}-\dfrac{6}{49}=\dfrac{159}{245}\)
b: \(\dfrac{4}{9}:\left(\dfrac{1}{15}-\dfrac{2}{3}\right)+\dfrac{4}{9}:\left(\dfrac{1}{11}-\dfrac{5}{22}\right)\)
\(=\dfrac{4}{9}:\left(\dfrac{1}{15}-\dfrac{10}{15}\right)+\dfrac{4}{9}:\left(\dfrac{2}{22}-\dfrac{5}{22}\right)\)
\(=\dfrac{4}{9}:\dfrac{-9}{15}-\dfrac{4}{9}:\dfrac{-3}{22}\)
\(=\dfrac{4}{9}\cdot\dfrac{-5}{3}+\dfrac{4}{9}\cdot\dfrac{22}{3}=\dfrac{-20+88}{27}=\dfrac{68}{27}\)