Bài 1. 

a) A = -x² - 4x - 2 = -( x² + 4x + 4 ) + 2 = -( x + 2 )² + 2

\(-\left(x+2\right)^2\le0\forall x\Rightarrow-\left(x+2\right)^2+2\le2\)

Đẳng thức xảy ra <=> x + 2 = 0 => x = -2

=> MaxA = 2 <=> x = -2

b) B = -2x² - 3x + 5 = -2( x² + 3/2x + 9/16 ) + 49/8 = -2( x + 3/4 )² + 49/8

\(-2\left(x+\frac{3}{4}\right)^2\le0\forall x\Rightarrow-2\left(x+\frac{3}{4}\right)^2+\frac{49}{8}\le\frac{49}{8}\)

Đẳng thức xảy ra <=> x + 3/4 = 0 => x = -3/4

=> MaxB = 49/8 <=> x = -3/4

c) C = ( 2 - x )( x + 4 ) = -x² - 2x + 8 = -( x² + 2x + 1 ) + 9 = -( x + 1 )² + 9

\(-\left(x+1\right)^2\le0\forall x\Rightarrow-\left(x+1\right)^2+9\le9\)

Đẳng thức xảy ra <=> x + 1 = 0 => x = -1

=> MaxC = 9 <=> x = -1

d) D = -8x² + 4xy - y² + 3 = -( 4x² - 4xy + y² ) - 4x² + 3 = -( 2x - y )² - 4x² + 3

\(\hept{\begin{cases}-\left(2x-y\right)^2\le0\forall x,y\\-4x^2\le0\forall x\end{cases}}\Rightarrow-\left(2x-y\right)^2-4x^2+3\le3\)

Đẳng thức xảy ra <=> \(\hept{\begin{cases}2x-y=0\\4x=0\end{cases}}\Rightarrow x=y=0\)

=> MaxD = 3 <=> x = y = 0

Bài 2.

a) A = x² - 2x + 5 = ( x² - 2x + 1 ) + 4 = ( x - 1 )² + 4

\(\left(x-1\right)^2\ge0\forall x\Rightarrow\left(x-1\right)^2+4\ge4\)

Đẳng thức xảy ra <=> x - 1 = 0 => x = 1

=> MinA = 4 <=> x = 1

b) B = x² - x + 1 = ( x² - 2.1/2.x + 1/4 ) + 3/4 = ( x - 1/2 )² + 3/4

\(\left(x-\frac{1}{2}\right)^2\ge0\forall x\Rightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

Đẳng thức xảy ra <=> x - 1/2 = 0 => x = 1/2

=> MinB = 3/4 <=> x = 1/2

c) C = ( x - 1 )( x + 2 )( x + 3 )( x + 6 )

C = [( x - 1 )( x + 6 )][( x + 2 )( x + 3)]

C = [ x² + 5x - 6 ][ x² + 5x + 6 ]

C = [ ( x² + 5x ) - 6 ][ ( x² + 5x ) + 6 ]

C = ( x² + 5x )² - 36

\(\left(x^2+5x\right)^2\ge0\forall x\Rightarrow\left(x^2+5x\right)^2-36\ge-36\)

Đẳng thức xảy ra <=> \(x^2+5x=0\Rightarrow x\left(x+5\right)=0\Rightarrow\orbr{\begin{cases}x=0\\x+5=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)

=> MinC = -36 <=> x = 0 hoặc x = -5

d) D = x² + 5y² - 2xy + 4y + 3 

D = ( x² - 2xy + y² ) + ( 4y² + 4y + 1 ) + 2

D = ( x - y )² + ( 2y + 1 )² + 2

\(\hept{\begin{cases}\left(x-y\right)^2\ge0\forall x,y\\\left(2y+1\right)^2\ge0\forall y\end{cases}}\Rightarrow\left(x-y\right)^2+\left(2y+1\right)^2+2\ge2\)

Đẳng thức xảy ra <=> \(\hept{\begin{cases}x-y=0\\2y+1=0\end{cases}\Rightarrow}x=y=-\frac{1}{2}\)

=> MinD = 2 <=> x = y = -1/2

a) ĐKXĐ : \(\hept{\begin{cases}x\ne0\\x\ne-2\end{cases}}\)

\(N=\frac{\left(x+2\right)^2}{x}.\left(1-\frac{x^2}{x+2}\right)-\frac{x^2+6x+4}{x}\)

\(N=\frac{\left(x+2\right)^2}{x}.\frac{x+2-x^2}{x+2}-\frac{x^2+6x+4}{x}\)

\(N=\frac{\left(x+2\right)\left(x+2-x^2\right)-x^2-6x-4}{x}\)

\(N=\frac{x^2+2x-x^3+2x+4-2x^2-x^2-6x-4}{x}\)

\(N=\frac{-x^3-2x^2-2x}{x}\)

\(N=\frac{-x\left(x^2+2x+2\right)}{x}\)

\(N=-\left(x^2+2x+2\right)\)

b) \(N=-\left(x^2+2x+2\right)\)

\(\Leftrightarrow N=-\left(x^2+2x+1+1\right)\)

\(\Leftrightarrow N=-\left(x+1\right)^2-1\le-1\)

Max N = -1 \(\Leftrightarrow x=-1\)

Vậy .......................

-3x( x + 2 )² + ( x + 3 )( x - 1 )( x + 1 ) - ( 2x - 3 )²

= -3x( x² + 4x + 4 ) + ( x + 3 )( x² - 1 ) - ( 4x² - 12x + 9 )

= -3x³ - 12x² - 12x + x³ - x + 3x² - 3 - 4x² + 12x - 9

= ( -3x³ + x³ ) + ( -12x² + 3x² - 4x² ) + ( -12x - x + 12x ) + ( -3 - 9 )

= -2x³ - 13x² - x - 12

WTF đăng một loạt vầy ai dám làm @@

Mấy bài này trong sách bài tập cx có bài mẫu

tự lật sách ra học ik , đăng 1 loạt ai giải cho chép zô hết

a) ĐKXĐ : \(x\ne1\)

\(\frac{2x+1}{x+1}\le1\)

\(\Leftrightarrow\frac{2x+1}{x+1}-1\le0\)

\(\Leftrightarrow\frac{2x+1-x-1}{x+1}\le0\)

\(\Leftrightarrow\frac{x}{x+1}\le0\)

+) \(\hept{\begin{cases}x\ge0\\x+1\le0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge0\\x\le-1\end{cases}}}\)( vô lí )

+) \(\hept{\begin{cases}x\le0\\x+1\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x\le0\\x\ge-1\end{cases}\Leftrightarrow}-1\le x\le0}\)

Vậy .........

b) ĐKXĐ : x khác 0

\(\Leftrightarrow\frac{3x-5}{x}\le2\)

\(\Leftrightarrow\frac{3x-5}{x}-2\le0\)

\(\Leftrightarrow\frac{x-5}{x}\le0\)

\(\Leftrightarrow0< x\le5\)

c) ĐKXĐ : x khác 5 ; 1

 \(\frac{x-2}{x-5}-\frac{3}{x-1}< 1\)

\(\Leftrightarrow\frac{\left(x-2\right)\left(x-1\right)-3\left(x-5\right)-\left(x-5\right)\left(x-1\right)}{\left(x-5\right)\left(x-1\right)}< 0\)

\(\Leftrightarrow\frac{x^2-x-2x+2-3x+15-\left(x^2-x-5x+5\right)}{\left(x-5\right)\left(x-1\right)}< 0\)

\(\Leftrightarrow\frac{x^2-6x+17-x^2+6x-5}{\left(x-5\right)\left(x-1\right)}< 0\)

\(\Leftrightarrow\frac{12}{\left(x-5\right)\left(x-1\right)}< 0\)

\(\Leftrightarrow\left(x-5\right)\left(x-1\right)< 0\)

\(\Leftrightarrow1< x< 5\)

d) ĐKXĐ : x khác 0

\(x+\frac{6}{x}< 7\)

\(\Leftrightarrow\frac{x^2-7x+6}{x}< 0\)

\(\Leftrightarrow\frac{x^2-7x+\frac{49}{4}-\frac{25}{4}}{x}< 0\)

\(\Leftrightarrow\frac{\left(x-\frac{7}{2}\right)^2-\frac{25}{4}}{x}< 0\)

\(\Leftrightarrow\frac{\left(x-6\right)\left(x-1\right)}{x}< 0\)

Bạn tự giải nốt ra nha  

cmr là gì?

A=(x-5)^2+(x+4)(1-x)

A=x^2 - 10x +25 + x - x^2 +4 - 4x

A= -13x +29

Vậy giá trị của biểu thức này phụ thuộc vào biến x

1. \(\frac{7x-1}{6}+2x=\frac{16-x}{5}\)

\(\Leftrightarrow5\left(7x-1\right)+60x=6\left(16-x\right)\)

\(\Leftrightarrow35x-5+60x=96-6x\)

\(\Leftrightarrow95x-5=96-6x\)

\(\Leftrightarrow95x+6x=96+5\)

\(\Leftrightarrow101x=101\)

\(\Leftrightarrow x=1\)

2. \(\frac{10x+3}{12}=1+\frac{6+8x}{9}\) 

\(\Leftrightarrow3\left(10x+3\right)=36+4\left(6+8x\right)\)

\(\Leftrightarrow30x+9=36+24+32x\)

\(\Leftrightarrow30x+9=32x+60\)

\(\Leftrightarrow30x-32x=60-9\)

\(\Leftrightarrow-2x=51\)

\(\Leftrightarrow x=-\frac{51}{2}\)

3. \(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)

\(\Leftrightarrow8x-3-2\left(3x-2\right)=2\left(2x-1\right)+x+3\)

\(\Leftrightarrow8x-3-6x+4=4x-2+x+3\)

\(\Leftrightarrow2x+1=5x+1\)

\(\Leftrightarrow2x=5x\)

\(\Leftrightarrow x=0\)

4) \(\frac{3\left(3-x\right)}{8}+\frac{2\left(5-x\right)}{3}=\frac{1-x}{2}-2\)

=> \(\frac{9-3x}{8}+\frac{10-2x}{3}=\frac{1-x}{2}-\frac{2}{1}\)

=> \(\frac{3\left(9-3x\right)}{24}+\frac{8\left(10-2x\right)}{24}=\frac{12\left(1-x\right)}{24}-\frac{48}{24}\)

=> \(\frac{27-9x}{24}+\frac{80-16x}{24}=\frac{12-12x}{24}-\frac{48}{24}\)

=> \(\frac{27-9x+80-16x}{24}=\frac{12-12x-48}{24}\)

=> 27 - 9x + 80 - 16x = 12 - 12x - 48

=> 27 - 9x + 80 - 16x - 12 + 12x + 48 = 0

=> (27 + 80 - 12 + 48) + (-9x - 16x + 12x) = 0

=> 143 - 13x = 0

=> 13x = 143

=> x = 11

5) \(\frac{2\left(x-3\right)}{7}+\frac{x-5}{3}-\frac{13x+4}{21}=0\)

=> \(\frac{2x-6}{7}+\frac{x-5}{3}-\frac{13x+4}{21}=0\)

=> \(\frac{3\left(2x-6\right)}{21}+\frac{7\left(x-5\right)}{21}-\frac{13x+4}{21}=0\)

=> \(\frac{6x-18}{21}+\frac{7x-35}{21}-\frac{13x+4}{21}=0\)

=> \(\frac{6x-18+7x-35-13x-4}{21}=0\)

=> 6x - 18 + 7x - 35 - 13x - 4 = 0

=> (6x + 7x - 13x) + (-18 - 35 - 4) = 0

=> -57 = 0(vô nghiệm)

6) \(\frac{6x+5}{2}-\left(2x+\frac{2x+1}{2}\right)=\frac{10x+3}{4}\)

=> \(\frac{6x+5}{2}-\frac{10x+3}{4}=2x+\frac{2x+1}{2}\)

=> \(\frac{2\left(6x+5\right)}{4}-\frac{10x+3}{4}=\frac{8x}{4}+\frac{2\left(2x+1\right)}{4}\)

=> \(\frac{12x+10}{4}-\frac{10x+3}{4}=\frac{8x}{4}+\frac{4x+2}{4}\)

=> \(\frac{12x+10-\left(10x+3\right)}{4}=\frac{8x+4x+2}{4}\)

=> \(\frac{12x+10-10x-3}{4}=\frac{12x+2}{4}\)

=> \(12x+10-10x-3=12x+2\)

=> \(2x+10-3=12x+2\)

=> 2x + 10 - 3 - 12x - 2 = 0

=> (2x - 12x) + (10 - 3 - 2) = 0

=> -10x + 5 = 0

=> -10x = -5

=> x = 1/2

7) \(\frac{2x-1}{5}-\frac{x-2}{3}-\frac{x+7}{15}=0\)

=> \(\frac{3\left(2x-1\right)}{15}-\frac{5\left(x-2\right)}{15}-\frac{x+7}{15}=0\)

=> \(\frac{6x-3}{15}-\frac{5x-10}{15}-\frac{x+7}{15}=0\)

=> \(\frac{6x-3-\left(5x-10\right)-\left(x+7\right)}{15}=0\)

=> 6x - 3 - 5x + 10 - x - 7 = 0

=> (6x - 5x - x) + (-3 + 10 - 7) = 0

=> 0x + 0 = 0

=> 0x = 0

=> x tùy ý

Bài 8 tự làm nhé