\(A+\frac{1}{4}=x^2+x+\frac{1}{4}=\left(x+\frac{1}{2}\right)^2\ge0\Rightarrow A_{min}=0-\frac{1}{4}=-\frac{1}{4}\Leftrightarrow x=-\frac{1}{2}\)

Ta có :

\(A=x^2+x\)

\(\Rightarrow A+\left(\frac{1}{2}\right)^2=x^2+x+\left(\frac{1}{2}\right)^2\)

\(\Leftrightarrow A+\frac{1}{4}=\left(x+\frac{1}{2}\right)^2\)

Có : \(\left(x+\frac{1}{2}\right)^2\ge0\Rightarrow\)\(A+\frac{1}{4}\ge0\Rightarrow A\ge-\frac{1}{4}\)

Vậy \(A_{min}=-\frac{1}{4}\)đạt được khi \(\left(x+\frac{1}{2}\right)^2=0\Leftrightarrow x+\frac{1}{2}=0\Leftrightarrow x=-\frac{1}{2}\)

_Vi hạ_

\(P=y\left(x^{n-1}+y^{n-1}\right)-x^{n-1}\left(x+y\right)\)

\(P=x^{n-1}y+y^n-x^n-x^{n-1}y\)

\(P=y^n-x^n\)

\(y\left(x^{n-1}+y^{n-1}\right)-x^{n-1}\left(x+y\right)=y^n+yx^{n-1}-x^n-yx^{n-1}=y^n-x^n\)

\(\frac{2x+5}{x+5}\in Z\Leftrightarrow2x+5⋮x+5\Leftrightarrow-5⋮x+5\Leftrightarrow x+5\in\left\{-1;1-5;5\right\}\Leftrightarrow x\in\left\{-6;-4;0;-10\right\}\)

TL:

\(\frac{2x+5}{x+5}=\frac{2x+10-5}{x+5}\)

\(=2-\frac{5}{x+5}\) 

Để BT đạt GT nguyên thì \(x+5\inƯ\left(5\right)\) 

\(\Rightarrow x+5\in\left\{\pm1;\pm5\right\}\) 

\(\Rightarrow x\in\left\{-4;-6;0;-10\right\}\) 

Vậy.........

6x² - ( 2x + 5 ).( 3x - 2 ) = 7

\(\Leftrightarrow\) 6x² - ( 6x² - 4x + 15x  - 10 ) = 7

\(\Leftrightarrow\) 6x² - ( 6x² + 11x - 10 ) = 7

\(\Leftrightarrow\) 6x² - 6x² - 11x + 10 = 7

\(\Leftrightarrow\) -11x = 7 -10

\(\Leftrightarrow\) - 11x = -3

\(\Leftrightarrow\)  \(x=\frac{3}{11}\)

Vậy x = \(\frac{3}{11}\)

6x² - (2x + 5)(3x - 2) = 7

<=> 6x² - 6x² - 11x + 10 = 7

<=> -11x + 10 = 7

<=> -11x = 7 - 10

<=> -11x = -3

<=> x = 3/11

=> x = 3/11

\(=>x^3+6x^2+12x+8-x^3+27+6x^2+12x+6=15\)

\(=>12x^2+24x+41-15=0\)

\(=>12x^2+24x+26=0\)

\(=>12\left(x^2+2x+1\right)+14=0\)

\(=>12\left(x+1\right)^2+14=0\)

\(=>2[6\left(x+1\right)^2+7]=0\)

\(=>6\left(x+1\right)^2+7=0\)

Mà \(\left(x+1\right)^2\ge0\)nên \(6\left(x+1\right)^2+7>0\)

Vậy ko có giá trị x nào thỏa mãn đề bài

( A - B )² = A² - 2AB + B²

A² + B² = ( A - B )² + 2AB

A² + B² =  12² + 10

A² + B² = 144 + 10 = 154

\(B=2x^2+2xy+5y^2-8x-22y\)

\(=\left(x^2+2xy+y^2\right)+\left(x^2-8x+16\right)+\left(4y^2-22y+\frac{484}{16}\right)-\frac{185}{4}\)

\(=\left(x+y\right)^2+\left(x-4\right)^2+\left(2y-\frac{22}{4}\right)^2-\frac{185}{4}\ge-\frac{185}{4}\)

Dấu = xảy ra khi :

.........................

Bn tự giải nốt nhé, mk ko bt có đúng hay ko , nếu sai thì thông cảm nha........

Mạn phép bỏ câu a :))

b) a²(b² - a²) + b²(b² + a²)

= a².b² + a².(-a²) + b².b² + b².a²

= a².b² - a⁴ + b⁴ + a².b² 

= a⁴ + 2a²b² + b² (hđt)

c) x²(x³ + 2y - x²y) - y(x² - x⁴ + y)

= x².x³ + x².2y + x².(-x²y) + (-y).x² + (-y).(-x)⁴ + (-y).y

= x⁵ + 2x²y - x⁴y - x²y + x⁴y - y² 

= x⁵ + (2xy² - xy²) + (-x⁴y + x⁴y) - y²

= x⁵ + xy² - y²