tìm GTNN của A=5x^2 - 25x +35 +7y^8
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Giups mk vs thank cacs bn
5 tháng 2 2021
b) PT \(\Leftrightarrow15x\left(5x+3\right)-35\left(5x+3\right)=0\)
\(\Leftrightarrow\left(15x-35\right)\left(5x+3\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=-\dfrac{3}{5}\end{matrix}\right.\)
Vậy \(S=\left\{-\dfrac{3}{5};\dfrac{7}{3}\right\}\)
c) PT \(\Leftrightarrow\left(2-3x\right)\left(x-11\right)+\left(2-3x\right)\left(2-5x\right)=0\)
\(\Leftrightarrow\left(2-3x\right)\left(-9-4x\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{9}{4}\end{matrix}\right.\)
Vậy \(S=\left\{\dfrac{2}{3};-\dfrac{9}{4}\right\}\)
R
5 tháng 2 2021
a)(x-1)(5x+3)=(3x-8)(x-1)
\(\Leftrightarrow\)(x-1)(5x+3)-(3x-8)(x-1)=0
\(\Leftrightarrow\left(x-1\right)\left(5x-3-3x+8\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x-5\right)=0\)
\(\left[{}\begin{matrix}x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{5}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{1;\dfrac{5}{2}\right\}\)
NH
1
TS
21 tháng 1 2016
a) (x - 1)(5x + 3) = (3x - 8)(x - 1)
\(\Leftrightarrow\left(x-1\right)\left(5x+3\right)-\left(3x-8\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(5x+3-3x+8\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+11\right)=0\)
\(\Leftrightarrow x-1=0\Rightarrow x=1\)
và\(2x+11=0\Rightarrow x=\frac{-11}{2}\)
LD
2
17 tháng 7 2017
\(=\sqrt{\left(5x-2\right)^2}+\sqrt{\left(5x\right)^2}\)= \(\left|2-5x\right|+\left|5x\right|\ge2+5x-5x=2\)
min A=2 \(\Leftrightarrow\hept{\begin{cases}2-5x\ge0\\5x\ge0\end{cases}\Leftrightarrow0\le x\le\frac{2}{5}}\)
NN
1
0
26 tháng 3 2018
a. (x−1)(5x+3)=(3x−8)(x−1)(x−1)(5x+3)=(3x−8)(x−1)
⇔(x−1)(5x+3)−(3x−8)(x−1)=0⇔(x−1)[(5x+3)−(3x−8)]=0⇔(x−1)(5x+3−3x+8)=0⇔(x−1)(2x+11)=0⇔(x−1)(5x+3)−(3x−8)(x−1)=0⇔(x−1)[(5x+3)−(3x−8)]=0⇔(x−1)(5x+3−3x+8)=0⇔(x−1)(2x+11)=0
⇔x−1=0⇔x−1=0hoặc 2x+11=02x+11=0
+ x−1=0⇔x=1x−1=0⇔x=1
+ 2x+11=0⇔x=−5,52x+11=0⇔x=−5,5
Phương trình có nghiệm x = 1 hoặc x = -5,5
b. 3x(25x+15)−35(5x+3)=03x(25x+15)−35(5x+3)=0
⇔15x(5x+3)−35(5x+3)=0⇔(15x−35)(5x+3)=0⇔15x(5x+3)−35(5x+3)=0⇔(15x−35)(5x+3)=0
⇔15x−35=0⇔15x−35=0 hoặc 5x+3=05x+3=0
+ 15x−35=0⇔x=3515=7315x−35=0⇔x=3515=\(\frac{7}{3}\)
+ 5x+3=0⇔x=−355x+3=0⇔x=−\(\frac{3}{5}\)
Phương trình có nghiệm x=\(\frac{7}{3}\)x=\(\frac{7}{3}\) hoặc x=−\(\frac{3}{5}\)
MH
6 tháng 2 2022
\(A=\dfrac{x^2+5x+8}{5}\)
\(=\dfrac{\left(x^2+5x+\dfrac{25}{4}\right)+\dfrac{7}{4}}{5}\)
\(=\dfrac{\left(x+\dfrac{5}{2}\right)^2}{5}+\dfrac{7}{20}\)
Vì \(\dfrac{\left(x+\dfrac{5}{2}\right)^2}{5}\ge0,\text{∀x}\)
⇒ \(A\ge\dfrac{7}{20},\text{∀x}\)
Min \(A=\dfrac{7}{20}\)⇔\(x=-\dfrac{5}{2}\)
6 tháng 2 2022
\(A=\dfrac{x^2+5x+8}{5}=\dfrac{\left(x^2+2.\dfrac{5}{2}x+\dfrac{25}{4}\right)+\dfrac{7}{4}}{5}=\dfrac{\left(x+\dfrac{5}{2}\right)^2+\dfrac{7}{4}}{5}\ge\dfrac{\dfrac{7}{4}}{5}=\dfrac{7}{4}.\dfrac{1}{5}=\dfrac{7}{20}\)-GTNN của A là \(\dfrac{7}{20}\Leftrightarrow x+\dfrac{5}{2}=0\Leftrightarrow x=\dfrac{-5}{2}\)
PT
1
CM
7 tháng 4 2018
a) x + 30 % x = − 1 , 3
x 1 + 3 10 = − 13 10 13 10 x = − 13 10 x = − 1
b) 1 3 x + 2 5 x − 1 = 0
1 3 x + 2 5 x − 2 5 = 0 11 15 x = 2 5 x = 2 5 : 11 15 x = 6 11
c) 3 x − 1 2 − 5 x + 3 5 = − x + 1 5
3 x − 3 2 − 5 x − 3 = − x + 1 5 x = − 3 2 − 3 − 1 5 x = − 47 10
\(A=5x^2-25x+35+7y^8\)
\(=5\left(x^2-5x+7\right)+7y^8\)
\(=5\left(x^2-5x+\frac{25}{4}+\frac{3}{4}\right)+7y^8\)
\(=5\left[\left(x-\frac{5}{2}\right)^2+\frac{3}{4}\right]+7y^8\)
\(=5\left(x-\frac{5}{2}\right)^2+\frac{15}{4}+7y^8\ge\frac{15}{4}\)
\(\Leftrightarrow x=\frac{5}{2};y=0\)