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
1) \(\dfrac{4^2}{2^3}=\dfrac{\left(2^2\right)^2}{2^3}=\dfrac{2^4}{2^3}=2\)
2) \(\dfrac{25^5}{125^3}=\dfrac{\left(5^2\right)^5}{\left(5^3\right)^3}=\dfrac{5^{10}}{5^9}=5\)
3) \(\dfrac{27^6}{9^9}=\dfrac{\left(3^3\right)^6}{\left(3^2\right)^9}=\dfrac{3^{18}}{3^{18}}=1\)
4) \(\dfrac{16^{13}}{32^{10}}=\dfrac{\left(2^4\right)^{13}}{\left(2^5\right)^{10}}=\dfrac{2^{52}}{2^{50}}=2^2-4\)
5) \(\dfrac{16^5}{64^4}=\dfrac{\left(4^2\right)^5}{\left(4^3\right)^4}=\dfrac{4^{10}}{4^{12}}=\dfrac{1}{4^2}=\dfrac{1}{16}\)
6) \(\dfrac{81^8}{27^{11}}=\dfrac{\left(3^4\right)^8}{\left(3^3\right)^{11}}=\dfrac{3^{32}}{3^{33}}=\dfrac{1}{3}\)
7) \(\dfrac{6^3}{2^3}=\dfrac{2^3\cdot3^3}{2^3}=3^3=27\)
8) \(\dfrac{5^4}{15^3}=\dfrac{5^4}{3^3\cdot5^3}=\dfrac{5}{3^3}=\dfrac{5}{27}\)
9) \(\dfrac{7^{15}}{14^{13}}=\dfrac{7^{15}}{7^{13}\cdot2^{13}}=\dfrac{7^2}{2^{13}}=\dfrac{49}{2^{13}}\)
10) \(\dfrac{\left(-2\right)^6}{24^2}=\dfrac{2^6}{8^2\cdot3^2}=\dfrac{2^6}{\left(2^3\right)^2\cdot3^2}=\dfrac{2^6}{2^6\cdot3^2}=\dfrac{1}{3^2}=\dfrac{1}{9}\)
11: \(\dfrac{27^2}{\left(-18\right)^3}=\dfrac{-3^6}{\left(3^2\cdot2\right)^3}=\dfrac{-3^6}{3^6\cdot2^3}=\dfrac{-1}{8}\)
12: \(\dfrac{\left(-10\right)^8}{8^3\cdot25^4}=\dfrac{2^8\cdot5^8}{2^6\cdot5^8}=2^2=4\)
13: \(\dfrac{4^4\cdot8^3}{16^4}=\dfrac{2^8\cdot2^9}{2^{16}}=2\)
14: \(\dfrac{5^7\cdot9^2}{15^5}=\dfrac{5^7\cdot3^4}{5^5\cdot3^5}=\dfrac{5^2}{3}=\dfrac{25}{3}\)
15: \(\dfrac{21^{13}}{49^6\cdot\left(-27\right)^4}=\dfrac{-7^{13}\cdot3^{13}}{7^{12}\cdot3^{12}}=-7\cdot3=-21\)
16: \(\dfrac{\left(-18\right)^{21}\cdot27^4}{81^{13}\cdot16^5}=\dfrac{-3^{42}\cdot2^{21}\cdot3^{12}}{3^{52}\cdot2^{20}}=\dfrac{-3^{54}}{3^{52}}\cdot2=-3^2\cdot2=-18\)
17: \(\dfrac{45^{14}\cdot8^2}{6^5\cdot125^4\cdot81^6}=\dfrac{3^{28}\cdot5^{14}\cdot2^6}{2^5\cdot3^5\cdot3^{24}\cdot5^{12}}=\dfrac{3^{28}}{3^{29}}\cdot\dfrac{5^{14}}{5^{12}}\cdot\dfrac{2^6}{2^5}=\dfrac{5^2\cdot2}{3}=\dfrac{50}{3}\)
18: \(\dfrac{11\cdot3^{22}\cdot3^7-9^{15}}{\left(2\cdot3^{14}\right)^2}=\dfrac{11\cdot3^{29}-3^{30}}{2^2\cdot3^{28}}=\dfrac{3^{29}\left(11-3\right)}{3^{28}\cdot2^2}=3\cdot\dfrac{8}{4}=3\cdot2=6\)
19: \(\dfrac{8^5\cdot\left(-5\right)^8+\left(-2\right)^5\cdot10^9}{16^4\cdot5^7+20^8}\)
\(=\dfrac{2^{15}\cdot5^8-2^{14}\cdot5^9}{2^{16}\cdot5^7+2^{16}\cdot5^8}=\dfrac{2^{14}\cdot5^8\cdot\left(2-5\right)}{2^{16}\cdot5^7\cdot\left(1+5\right)}=\dfrac{1}{4}\cdot5\cdot\dfrac{-3}{6}=\dfrac{5}{4}\cdot\dfrac{-1}{2}=-\dfrac{5}{8}\)