\(\left[36x4x\left(82-7x11\right)^2\right]:4-2016^0\)

\(=\left[144x\left(82-77\right)^2\right]:4-1\)

\(=\left(144x5^2\right):4-1\)

\(=\left(144x25\right):4-1\)

\(=3600:4-1\)

\(=900-1\)

\(=899\)

[ 36.4.( 82 - 7.11 )² ] : 4 - 2016⁰ 

= [ 36.4.5² ] : 4 - 2016⁰

= [ 36.4.25 ] : 4 - 2016⁰

= 3600 : 4 -1

= 900 - 1 

= 899

Lời giải:

$x^2+4x-8y-2xy+2y^2+2024$

$=(x^2-2xy+y^2)+4x-8y+y^2+2024$

$=(x-y)^2+4(x-y)+y^2-4y+2024$

$=(x-y)^2+4(x-y)+4+(y^2-4y+4)+2016$

$=(x-y+2)^2+(y-2)^2+2016\geq 2016$

Vậy GTNN của biểu thức là 2016. Giá trị này đạt tại $x-y+2=y-2=0$

$\Leftrightarrow y=2; x=0$

Số lượng số hạng là:

(2024 - 0) : 2 + 1 = 1013 (số hạng)

Tổng dãy số trên là:

(2024 + 0) x 1013 : 2 = 1025156 

Tổng dãy \(0;2;4;6;8;...;2024\) là :

\(\left[\left(2024-0\right):2+1\right]x\left(0+2024\right):2\)

\(=1013x2024:2\)

\(=1025156\)

Đáp án : 1025156

\(\left(x-2\right)^4+\left(2y-1\right)^{2024}\le0\left(1\right)\)

Vì \(\left\{{}\begin{matrix}\left(x-2\right)^4\ge0\forall x\\\left(2y-1\right)^{2024}\ge0\forall x\end{matrix}\right.\)

\(\Rightarrow\left(x-2\right)^4+\left(2y-1\right)^{2024}\ge0\left(2\right)\)

Từ (1) và (2)

\(\Rightarrow\left(x-2\right)^4+\left(2y-1\right)^{2024}=0\)

\(\Rightarrow\left\{{}\begin{matrix}x=2\\y=\dfrac{1}{2}\end{matrix}\right.\)

\(M=21.2^2.\dfrac{1}{2}+4.2.\left(\dfrac{1}{2}\right)^2=21.2+4.2.\dfrac{1}{4}=42+2=44\)

Ta có: \(\left(x-2\right)^4\ge0\forall x\)

           \(\left(2y-1\right)^{2024}\ge0\forall y\)

\(\Rightarrow\left(x-2\right)^4+\left(2y-1\right)^{2024}\ge0\forall x;y\)

Mặt khác: \(\left(x-2\right)^4+\left(2y-1\right)^{2024}\le0\)

nên \(\left(x-2\right)^4+\left(2y-1\right)^{2024}=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-2\right)^4=0\\\left(2y-1\right)^{2024}=0\end{matrix}\right.\)

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

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

Thay \(x=2\) và \(y=\dfrac{1}{2}\) vào \(M\), ta được:

\(M=21\cdot2^2\cdot\dfrac{1}{2}+4\cdot2\cdot\left(\dfrac{1}{2}\right)^2\)

\(=42+2\)

\(=44\)

Vậy \(M=44\) tại \(x=2;y=\dfrac{1}{2}\).

#\(Toru\)

a) x⁴ + 2x³ + 5x² + 4x - 12 = 0

=> x⁴ - x³+ 3x³- 3x² + 8x² -8x + 12x - 12 = 0

=> x³( x - 1) + 3x²( x - 1) + 8x( x - 1) + 12 ( x - 1 ) = 0

=> ( x - 1)( x³ + 3x² + 8x + 12 ) = 0

=> ( x - 1)( x³ + 2x² + x² + 2x + 6x + 12 ) = 0

=> ( x - 1)[ x²( x + 2) + x( x + 2) + 6( x + 2) ] = 0

=> ( x - 1)( x + 2)( x² + x + 6 ) = 0

Ta thấy : x² + x + 6

= x² + 2.\(\dfrac{1}{2}x+\dfrac{1}{4}-\dfrac{1}{4}+6=\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{4}\text{≥}\dfrac{23}{4}>0\text{∀}x\)

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

=> x = 1 hoặc x = -2

Vậy,....

b) ( x + 1)³ + ( x - 2)³ = ( 2x - 1)³

=>x³+ 3x² + 3x + 1 + x³ - 6x² + 12x - 8 - ( 8x³ - 12x² + 6x - 1)=0

=> 2x³ - 3x² + 15x - 7 - 8x³ + 12x² - 6x + 1 = 0

=> 9x² - 6x³ + 9x - 6 = 0

=> 9x( x + 1) -6( x³ + 1 ) = 0

=> 9x( x + 1) - 6( x + 1)( x² - x + 1) = 0

=> 3( x + 1)( 3x - 2x² + 2x - 2) = 0

=> 3( x + 1)( - 2x² + 5x - 2) = 0

=> 3( x + 1)( - 2x² + x + 4x - 2) = 0

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

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

Suy ra :

* x + 1 = 0 => x = -1

* 1 - 2x = 0 => x = \(\dfrac{1}{2}\)

* x - 2 = 0 => x = 2

Vậy,.....

\(A=\dfrac{2024}{x^2-2x+8}\)

Ta có :

\(x^2-2x+8=x^2-2x+1+7=\left(x-1\right)^2+7\ge7,\forall x\inℝ\)

\(\Rightarrow\dfrac{1}{x^2-2x+8}\le\dfrac{1}{7}\)

\(\Rightarrow A=\dfrac{2024}{x^2-2x+8}\le\dfrac{2024}{7}\)

\(\Rightarrow Max\left(A\right)=\dfrac{2024}{7}\left(tạix=1\right)\)

Bài 2: 

a: \(x^2\left(x^2-16\right)=0\)

\(\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\)

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

b: \(x^8+36x^4=0\)

\(\Leftrightarrow x^4=0\)

hay x=0

a(b+3)-b(3+b)

=(3+b)(a-b)

Thay số, có: (3+1997).(2003-1997)

= 2000.6 =12000

xy(x+y)-2x-2y

xy(x+y)- 2(x+y)

(x+y).(xy-2)

Thay số, co: 7. (8-2)

7.4=28