1) \(f\left(x\right)=-3x^2-12x+5\)

\(\Rightarrow f\left(x\right)=-3\left(x^2+4x\right)+5\)

\(\Rightarrow f\left(x\right)=-3\left(x^2+4x+4\right)+5+12\)

\(\Rightarrow f\left(x\right)=-3\left(x+2\right)^2+17\le17\left(-3\left(x+2\right)^2\le0,\forall x\right)\)

\(\Rightarrow GTLN\left(f\left(x\right)\right)=17\left(tạix=-2\right)\)

2) \(f\left(x\right)=-8x^2+20x\)\

\(\Rightarrow f\left(x\right)=-8\left(x^2+\dfrac{5}{2}x\right)\)

\(\Rightarrow f\left(x\right)=-8\left(x^2+\dfrac{5}{2}x+\dfrac{25}{16}\right)+\dfrac{25}{2}\)

\(\Rightarrow f\left(x\right)=-8\left(x+\dfrac{5}{4}\right)^2+\dfrac{25}{2}\le\dfrac{25}{2}\left(-8\left(x+\dfrac{5}{4}\right)^2\le0,\forall x\right)\)

\(\Rightarrow GTLN\left(f\left(x\right)\right)=\dfrac{25}{2}\left(tạix=-\dfrac{5}{4}\right)\)

1) \(f\left(x\right)=6x^2-15x+4\)

\(\Rightarrow f\left(x\right)=6\left(x^2-\dfrac{5}{3}x\right)+4\)

\(\Rightarrow f\left(x\right)=6\left(x^2-\dfrac{5}{3}x+\dfrac{25}{36}-\dfrac{25}{36}\right)+4\)

\(\Rightarrow f\left(x\right)=6\left(x^2-\dfrac{5}{3}x+\dfrac{25}{36}\right)+4-\dfrac{25}{6}\)

\(\Rightarrow f\left(x\right)=6\left(x-\dfrac{5}{6}\right)^2-\dfrac{1}{6}\ge-\dfrac{1}{6}\left(6\left(x-\dfrac{5}{6}\right)^2\ge0,\forall x\right)\)

\(\Rightarrow GTNN\left(f\left(x\right)\right)=-\dfrac{1}{6}\left(tạix=\dfrac{5}{6}\right)\)

2) \(f\left(x\right)=4x^2-13x+5\)

\(\Rightarrow f\left(x\right)=4\left(x^2-\dfrac{13}{4}x\right)+5\)

\(\Rightarrow f\left(x\right)=4\left(x^2-\dfrac{13}{4}x+\dfrac{169}{64}-\dfrac{169}{64}\right)+5\)

\(\Rightarrow f\left(x\right)=4\left(x^2-\dfrac{13}{4}x+\dfrac{169}{64}\right)+5-\dfrac{169}{16}\)

\(\Rightarrow f\left(x\right)=4\left(x-\dfrac{13}{8}\right)^2-\dfrac{89}{16}\ge-\dfrac{89}{16}\left(4\left(x-\dfrac{13}{8}\right)^2\ge0,\forall x\right)\)

\(\Rightarrow GTNN\left(f\left(x\right)\right)=-\dfrac{89}{16}\left(tạix=\dfrac{13}{8}\right)\)

A=   (\(x-y\))² - (y-z)²

A = (\(x-y\) - y + z)(\(x-y\) + y -  z)

A = ( \(x\) - 2y + z)(\(x-z\))

(x+y)³-x(x+y)²=(x+y)²(x+y-x)=(x+y)²y

cám ơn bạn nha!!!

Ta có :

\(x-y=10\)

\(\Rightarrow\left(x-y\right)^2=100\left(x>y\right)\)

\(\Rightarrow\left(x+y\right)^2-4xy=100\)

\(\Rightarrow\left(x+y\right)^2=100+4xy\)

mà \(x.y=24\)

\(\Rightarrow\left(x+y\right)^2=100+4.24=196\)

\(\Rightarrow\left(x+y\right)^2=14^2\)

\(\Rightarrow\left[{}\begin{matrix}x+y=4\\x+y=-4\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}D=x+y=4\\D=x+y=-4\end{matrix}\right.\)

Đính Chính

\(x+y=\pm14\)

Anh đang trên xe đi chơi nên xin phép gõ không latex

--

(2x+1)^2 - 4x^2 + 4x -1 =0

<=> (2x+1)^2 - (2x-1)^2=0

<=> (2x + 1 + 2x -1). (2x+1 - 2x +1)=0

<=> 4x. 2= 0

<=> 8x=0

<=> x =0

`@` `\text {Ans}`

`\downarrow`

`(2x + 1)^2 - 4x^2 + 4x - 1 = 0`

`<=> 4x^2 + 4x + 1 - 4x^2 + 4x - 1 = 0`

`<=> (4x^2 - 4x^2) + (4x + 4x) + (1 - 1) = 0`

`<=> 8x = 0`

`<=> x = 0`

Vậy, `x = 0.`

\(\left(4x-1\right)^2-4\left(2x+1\right)^2-x-4=0\)

\(\Leftrightarrow\left(16x^2-8x+1\right)-4\left(4x^2+4x+1\right)-x-4=0\)

\(\Leftrightarrow16x^2-8x+1-16x^2-16x-4-x-4=0\)

\(\Leftrightarrow25x-7=0\)

\(\Leftrightarrow25x=7\)

\(\Leftrightarrow x=\dfrac{7}{25}\)

`@` `\text {Ans}`

`\downarrow`

`(4x - 1)^2 - 4(2x + 1)^2 - x - 4 = 0`

`<=> 16x^2 - 8x + 1 - 4(4x^2 + 4x + 1) - x - 4 = 0`

`<=> 16x^2 - 8x + 1 - 16x^2 - 16x - 4 - x - 4 = 0`

`<=> -25x - 7 = 0`

`<=> -25x = 7`

`<=> x =`\(\dfrac{-7}{25}\)

Vậy, \(x= \dfrac{-7}{25}\)

mik cần gấp giúp vs ạ

\(C=4x^2+y^2-4x+8y+12\)

\(C=4x^2-4x+1+y^2+8y+16-5\)
\(C=\left(4x^2-4x+1\right)+\left(y^2+8y+16\right)-5\)

\(C=\left(2x-1\right)^2+\left(y+4\right)^2-5\)

Mà: \(\left\{{}\begin{matrix}\left(2x-1\right)^2\ge0\forall x\\\left(y+4\right)^2\ge0\forall x\end{matrix}\right.\)

Nên: \(C=\left(2x-1\right)^2+\left(y+4\right)^2-5\ge-5\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}2x-1=0\\y+4=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-4\end{matrix}\right.\)

Vậy: \(C_{min}=-5\) khi \(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-4\end{matrix}\right.\)

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