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19 tháng 3 2021

GTNN=10 Khi x=9 y=16

19 tháng 3 2021

bạn làm ntn?

27 tháng 10 2019

a) \(A=\sqrt{x^2-2x+1}+\sqrt{x^2-6x+9}\)

\(=\sqrt{\left(x-1\right)^2}+\sqrt{\left(x-3\right)^2}\)

\(=\left|x-1\right|+\left|x-3\right|\ge\left|\left(x-1\right)+\left(3-x\right)\right|=2\)

Vậy\(A_{min}=2\Leftrightarrow\left(x-1\right)\left(3-x\right)\ge0\)

\(TH1:\hept{\begin{cases}x-1\ge0\\3-x\ge0\end{cases}}\Leftrightarrow1\le x\le3\)

\(TH1:\hept{\begin{cases}x-1\le0\\3-x\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le1\\x\ge3\end{cases}}\left(L\right)\)

Vậy \(A_{min}=2\Leftrightarrow1\le x\le3\)

2:

a: A(x)=0

=>5x-10-2x-6=0

=>3x-16=0

=>x=16/3

b: B(x)=0

=>5x^2-125=0

=>x^2-25=0

=>x=5 hoặc x=-5

c: C(x)=0

=>2x^2-x-3=0

=>2x^2-3x+2x-3=0

=>(2x-3)(x+1)=0

=>x=3/2 hoặc x=-1

26 tháng 12 2022

\(B1\\ a,2x+10y=2\left(x+5y\right)\\ b,x^2+4x+4=x^2+2.2x+2^2=\left(x+2\right)^2\\ c,x^2-y^2+10y-25\\ =\left(x^2-y^2\right)+5\left(2y-5\right)\\ =\left(x-y\right)\left(x+y\right)+5\left(2y-5\right)\\ B2\)

\(a,x^2-3x+x-3=0\\ =>x\left(x-3\right)+\left(x-3\right)=0\\ =>\left(x+1\right)\left(x-3\right)=0\\ =>\left[{}\begin{matrix}x+1=0\\x-3=0\end{matrix}\right.=>\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\\ b,2x\left(x-3\right)-\dfrac{1}{2}\left(4x^2-3\right)=0\\ =>2x^2-6x-2x^2+\dfrac{3}{2}=0\\ =>-6x=-\dfrac{3}{2}\\ =>x=\left(-\dfrac{3}{2}\right):\left(-6\right)\\ =>x=\dfrac{1}{4}\\ c,x^2-\left(x-3\right)\left(2x-5\right)=9\\ =>x^2-2x^2+6x+5x-15=9\\ =>-x^2+11-15-9=0\\ =>-x^2+11x-24=0\\ =>-x^2+8x+3x-24=0\\ =>-x\left(x-8\right)+3\left(x-8\right)=0\\ =>\left(3-x\right)\left(x-8\right)=0\\ =>\left[{}\begin{matrix}3-x=0\\x-8=0\end{matrix}\right.=>\left[{}\begin{matrix}x=3\\x=8\end{matrix}\right.\)

15 tháng 5 2016

Toán lớp 9

5 tháng 9 2021

a. (x - 22) - 1 = 0

<=> x - 4 - 1 = 0

<=> x = 5

b. 4 - (x - 2)2 = 0

<=> 22 - (x - 2)2 = 0

<=> (2 - x + 2)(2 + x - 2) = 0

<=> x(4 - x) = 0

<=> \(\left[{}\begin{matrix}x=0\\4-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

 

5 tháng 9 2021

d. (3x - 2)2 - (2x + 3)2 = 5(x + 4)(x - 4)

<=> (3x - 2 - 2x - 3)(3x - 2 + 2x + 3) = 5(x2 - 16)

<=> (x - 5)(5x + 1) = 5x2 - 80

<=> 5x2 + x - 25x - 5 = 5x2 - 80

<=> 5x2 - 5x2 + x - 25x = -80 + 5

<=> -24x = -75

<=> x = \(\dfrac{25}{8}\)

28 tháng 10 2019

1 M=\(x^2-4xy+4y^2-2x+4y+10\)

=\(\left(x^2-4xy+4y^2\right)+\left(-2x+4y\right)+10\)

\(=\left(x-2y\right)^2-2\left(x-2y\right)+10\)

\(=\left(x-2y\right)\left(x-2y-2\right)+10\)

\(\left(x-2y\right)\left(x-2y-2\right)\ge0\)

nên \(\left(x-2y\right)\left(x-2y-2\right)+10\ge10\)

\(\Rightarrow\)A\(\ge13\)

dấu "=" xảy ra khi (x-2y)(x-2y-2)=0

\(\left[{}\begin{matrix}x-2y=0\\x-2y-2=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}2y=x\\x-2y=2\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=0;y=0\\x=2;y=1\end{matrix}\right.\)

vậy GTNN của M=10 khi x=0; y=0

x=2;y=1

15 tháng 10 2023

a: 78x(x-97)-x+97=0

=>(x-97)(78x-1)=0

=>\(\left[{}\begin{matrix}x=97\\x=\dfrac{1}{78}\end{matrix}\right.\)

b: \(\dfrac{2}{3}x\left(x^2-4\right)=0\)

=>\(x\left(x-2\right)\left(x+2\right)=0\)

=>\(\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)

c: \(\left(x+2\right)^2-\left(x-2\right)\left(x+2\right)=0\)

=>\(x^2+4x+4-x^2+4=0\)

=>4x+8=0

=>x+2=0

=>x=-2

15 tháng 10 2023

\(a,78x\left(x-97\right)-x+97=0\)

\(\Leftrightarrow78x\left(x-97\right)-\left(x-97\right)=0\)

\(\Leftrightarrow\left(x-97\right)\left(78x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-97=0\\78x-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=97\\x=\dfrac{1}{78}\end{matrix}\right.\)

\(b,\dfrac{2}{3}x\left(x^2-4\right)=0\)

\(\Leftrightarrow\dfrac{2}{3}x\left(x-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{3}x=0\\x-2=0\\x+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)

\(c,\left(x+2\right)^2-\left(x-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left[\left(x+2\right)-\left(x-2\right)\right]=0\)

\(\Leftrightarrow\left(x+2\right)\left(x+2-x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\cdot4=0\)

\(\Leftrightarrow x+2=0\)

\(\Leftrightarrow x=-2\)