làm nốt

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

=(6x²+x-2)(3-x)

=-6x³+17x²+5x-6

e) (x+3)(x²+3x-5)

=x³+6x²+4x-15

f) (xy-2)(x³-2x-6)

=x⁴y-2x³-2x²y-6xy+4x+12

g) (5x³-x2+2x-3)(4x²-x+2)

=20x⁵-9x⁴+19x³-16x²+7x-6

Bài 1:

a) (x-2)(x2+3x+4)

=x(5x+4)-2(5x+4)

= 5x²+4x-10x-8

=5x²-6x-8

a ) \(x^2.\frac{y^3}{5}=\frac{A}{35.\left(x+y\right)}\)

\(\Leftrightarrow5A=35.x^2.y^3.\left(x+y\right)\)

\(\Leftrightarrow A=7x^2y^3\left(x+y\right)\)

b ) \(\frac{x^2-4x+4}{x^2-4}=\frac{x-2}{A}\)

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

\(\Leftrightarrow A=\frac{\left(x-2\right)^2\left(x+2\right)}{\left(x-2\right)^2}=x+2\).

1/ Ta có : \(P\left(x\right)=-x^2+13x+2012=-\left(x-\frac{13}{2}\right)^2+\frac{8217}{4}\le\frac{8217}{4}\)

Dấu "=" xảy ra khi x = 13/2

Vậy Max P(x) = 8217/4 tại x = 13/2

2/ Ta có : \(x^3+3xy+y^3=x^3+3xy.1+y^3=x^3+y^3+3xy\left(x+y\right)=\left(x+y\right)^3=1\)

3/ \(a+b+c=0\Leftrightarrow\left(a+b+c\right)^2=0\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=0\)

\(\Leftrightarrow ab+bc+ac=-\frac{1}{2}\) \(\Leftrightarrow\left(ab+bc+ac\right)^2=\frac{1}{4}\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=\frac{1}{4}\)

\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2=\frac{1}{4}\)(vì a+b+c=0)

Ta có : \(a^2+b^2+c^2=1\Leftrightarrow\left(a^2+b^2+c^2\right)^2=1\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=1\)

\(\Leftrightarrow a^4+b^4+c^4=1-2\left(a^2b^2+b^2c^2+c^2a^2\right)=1-\frac{2.1}{4}=\frac{1}{2}\)

Sửa đề: Cho x,y,z đôi một khác nhau và \(x^3+y^3+z^3=3xyz\)

Ta có: \(x^3+y^3+z^3=3xyz\)

=>\(\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz=0\)

=>\(\left(x+y+z\right)\left\lbrack\left(x+y\right)^2-z\left(x+y\right)+z^2\right\rbrack-3xy\left(x+y+z\right)=0\)

=>\(\left(x+y+z\right)\left\lbrack x^2+2xy+y^2-xz-zy+z^2\right\rbrack-3xy\left(x+y+z\right)=0\)

=>\(\left(x+y+z\right)\left\lbrack x^2+y^2+z^2-xy-xz-yz\right\rbrack=0\)

=>\(\left(x+y+z\right)\left\lbrack2x^2+2y^2+2z^2-2xy-2yz-2xz\right\rbrack=0\)

=>\(\left(x+y+z\right)\left\lbrack\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2\right\rbrack=0\)

mà \(\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2>0\) vì x,y,z đôi một khác nhau

nên x+y+z=0

=>y+z=-x

Sửa đề: \(A=2025+\left(y+z\right)^{2025}+x^{2025}\)

\(=2025+\left(-x\right)_{}^{2025}+x^{2025}\)

\(=2025-x^{2025}+x^{2025}=2025\)

Bài 1:

a. A = x^2 - 5x - 1

\(=x^2-5x+\frac{25}{4}-\frac{29}{4}\)

\(=x^2-5x+\left(\frac{5}{2}\right)^2-\frac{29}{4}\)

\(=\left(x-\frac{5}{2}\right)^2-\frac{29}{4}\ge0-\frac{29}{4}=-\frac{29}{4}\)

Dấu = khi x=5/2

Vậy MinC=-29/4 khi x=5/2

2. Tìm x:
a. ( 2x - 3 )^2 - ( 4x + 1 )( 4x - 1 ) = ( 2x - 1 ).( 3 - 7x )

=>4x²-12x+9+1-16x²=-14x²+13x-3

=>-12x²-12x+10=-14x²+13x-3

=>2x²-25x+13=0

\(\Rightarrow2\left(x-\frac{25}{4}\right)^2-\frac{521}{8}=0\)

\(\Rightarrow\left(x-\frac{25}{4}\right)^2=\frac{521}{16}\)

\(\Rightarrow x-\frac{25}{4}=\pm\sqrt{\frac{521}{16}}\)

\(\Rightarrow x=\frac{25}{4}\pm\frac{\sqrt{521}}{4}\)

c. 4.( x - 3 ) - ( x + 2 ) = 0

=>4x-12-x-2=0

=>3x-14=0

=>3x=14

=>x=14/3

Bài 1: Giả sử \(C\ge0\)

Ta có:

\(C=b^3-a^3-6b^2-a^2+9b\ge0\)

\(\Leftrightarrow\left(b^3-6b^2+9b\right)-\left(a^3+a^2\right)\ge0\Leftrightarrow b\left(b^2-6b+9\right)-a^2\left(a+1\right)\ge0\)

\(\Leftrightarrow b\left(b-3\right)^2-a^2\left(a+1\right)\ge0\)

Mà \(a+b=3\Rightarrow b=3-a\)

\(\Rightarrow C=\left(3-a\right)\left(3-a-3\right)^2-a^2\left(a+1\right)\ge0\Leftrightarrow a^2\left(3-a\right)-a^2\left(a+1\right)=a^2\left(2-2a\right)\ge0\)

Ta có: \(a^2\ge0;a\le0\Rightarrow2a\le0\Rightarrow-2a\ge0\Rightarrow2-2a\ge2\Rightarrow C\ge0\)(luôn đúng)

Bài 2: để suy nghĩ đã á

nhanh len

Bài 2:

Ta có: \(\left(2x-1\right)^4\ge0\forall x\)

=>\(-\left(2x-1\right)^4\le0\forall x\)

=>\(A=-\left(2x-1\right)^4+5\le5\forall x\)

Dấu '=' xảy ra khi 2x-1=0

=>2x=1

=>\(x=\frac12\)

Bài 1:

a: \(x^4\ge0\forall x\)

\(\left(y-\frac27\right)^6\ge0\forall y\)

Do đó: \(x^4+\left(y-\frac27\right)^6\ge0\forall x,y\)

=>\(x^4+\left(y-\frac27\right)^6-2019\ge-2019\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}x=0\\ y-\frac27=0\end{cases}\Rightarrow\begin{cases}x=0\\ y=\frac27\end{cases}\)

b: \(\left(x-5\right)^2\ge0\forall x\)

\(\left|y-7\right|\ge0\forall y\)

Do đó: \(\left(x-5\right)^2+\left|y-7\right|\ge0\forall x,y\)

=>\(\left(x-5\right)^2+\left|y-7\right|+2000\ge2000\forall x,y\)

Dấu '=' xảy ra khi x-5=0 và y-7=0

=>x=5 và y=7

a). -121

b). Casio hoặc Phan Đăng Nhật Minh 

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