K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
MN
20 tháng 2 2021
\(11x^2-15x+4=0\)
\(\Leftrightarrow11x^2-11x-4x+4=0\)
\(\Leftrightarrow11x\left(x-1\right)-4\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(11x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\11x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{4}{11}\end{matrix}\right.\)
\(S=\left\{1,\dfrac{4}{11}\right\}\)
20 tháng 2 2021
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 \(\dfrac{4}{11}\)
HR
2
31 tháng 3 2021
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
HQ
2
13 tháng 12 2020
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!
13 tháng 12 2020
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 \(P=\frac{2}{3}\)
NT
1
a: \(\dfrac{-3}{4}+\left(3-\dfrac{1}{4}\right)-\left(2,25-\dfrac{9}{4}\right)\)
\(=-\dfrac{3}{4}+3-\dfrac{1}{4}-\left(2,25-2,25\right)\)
\(=-1+3=2\)
b: \(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{23}+\dfrac{1}{6}\)
\(=\dfrac{1}{2}+\dfrac{1}{6}-\dfrac{1}{3}+\dfrac{1}{23}\)
\(=\dfrac{3+1-2}{6}+\dfrac{1}{23}\)
\(=\dfrac{1}{3}+\dfrac{1}{23}=\dfrac{26}{69}\)
c: \(\left(-\dfrac{13}{7}-\dfrac{4}{9}\right)-\left(-\dfrac{10}{7}-\dfrac{4}{9}\right)\)
\(=-\dfrac{13}{7}-\dfrac{4}{9}+\dfrac{10}{7}+\dfrac{4}{9}\)
\(=-\dfrac{13}{7}+\dfrac{10}{7}=-\dfrac{3}{7}\)
d: \(\dfrac{-14}{12}+0,65-\left(-\dfrac{7}{42}-0,35\right)\)
\(=-\dfrac{7}{6}+0,65+0,35+\dfrac{7}{42}\)
\(=\dfrac{-49}{42}+\dfrac{7}{42}+1=-\dfrac{42}{42}+1=0\)
e: \(\left(\dfrac{7}{8}-\dfrac{5}{2}+\dfrac{4}{7}\right)-\left(-\dfrac{3}{7}+1-\dfrac{13}{8}\right)\)
\(=\dfrac{7}{8}-\dfrac{5}{2}+\dfrac{4}{7}+\dfrac{3}{7}-1+\dfrac{13}{8}\)
\(=\dfrac{20}{8}-\dfrac{5}{2}=\dfrac{5}{2}-\dfrac{5}{2}=0\)
f: \(\dfrac{-3}{7}+\left(3-\dfrac{3}{4}\right)-\left(2,25-\dfrac{10}{7}\right)\)
\(=-\dfrac{3}{7}+2,25-2,25+\dfrac{10}{7}\)
\(=\dfrac{10}{7}-\dfrac{3}{7}=\dfrac{7}{7}=1\)
g: \(\dfrac{1}{2}-\dfrac{43}{101}+\left(-\dfrac{1}{3}\right)-\dfrac{1}{6}\)
\(=\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)-\dfrac{43}{101}\)
\(=\left(\dfrac{3}{6}-\dfrac{2}{6}-\dfrac{1}{6}\right)-\dfrac{43}{101}=0-\dfrac{43}{101}=-\dfrac{43}{101}\)
h: \(\left(\dfrac{5}{3}-\dfrac{3}{7}+9\right)-\left(2+\dfrac{5}{7}-\dfrac{2}{3}\right)+\left(\dfrac{8}{7}-\dfrac{4}{3}-10\right)\)
\(=\dfrac{5}{3}-\dfrac{3}{7}+9-2-\dfrac{5}{7}+\dfrac{2}{3}+\dfrac{8}{7}-\dfrac{4}{3}-10\)
\(=\left(\dfrac{5}{3}+\dfrac{2}{3}-\dfrac{4}{3}\right)+\left(-\dfrac{3}{7}-\dfrac{5}{7}+\dfrac{8}{7}\right)+\left(9-2-10\right)\)
\(=\dfrac{2}{3}-3=-\dfrac{7}{3}\)
i: \(\dfrac{1}{2}+\dfrac{5}{6}-\dfrac{1}{3}=\dfrac{3}{6}+\dfrac{5}{6}-\dfrac{2}{6}=\dfrac{5+1}{6}=\dfrac{6}{6}=1\)
k: \(\dfrac{1}{2}-\left[\dfrac{3}{8}+\left(-\dfrac{7}{4}\right)\right]\)
\(=\dfrac{1}{2}-\dfrac{3}{8}+\dfrac{7}{4}\)
\(=\dfrac{4}{8}-\dfrac{3}{8}+\dfrac{14}{8}=\dfrac{15}{8}\)
mình cảm ơn nha