Tìm giá trị lớn nhất của cos √x + π/6

\(A=\left(a+b+c\right)^3-\left(b+c-a\right)^3-\left(c+a-b\right)^3-\left(a+b-c\right)^3\)

Đặt \(x=b+c-a,y=c+a-b,z=a+b-c\).

Khi đó \(x+y+z=a+b+c\).

\(A=\left(x+y+z\right)^3-x^3-y^3-z^3\)

\(=\left(y+z\right)\left[\left(x+y+z\right)^2+x\left(x+y+z\right)+x^2\right]-\left(y+z\right)\left(y^2-yz+z^2\right)\)

\(=\left(y+z\right)\left(3x^2+3xy+3zx+3yz\right)\)

\(=3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)

\(=24abc=24\)

Đặt \(\hept{\begin{cases}a+b-c=x\\b+c-a=y\\c+a-b=n\end{cases}}\)

\(\Rightarrow x+y+n=a+b+c\)

\(\Rightarrow x+y+2b\)

\(\Rightarrow y+n=2c\)

\(\Rightarrow n+x=2a\)

Ta có:

\(A=\left(x+y+n\right)^3-y^3-n^3-x^3\)

\(=x^3+y^3+n^3+3\left(x+y\right)\left(y+n\right)\left(n+x\right)-y^3-n^3-x^3\)

\(=3\left(x+y\right)\left(y+n\right)\left(n+z\right)\)

\(=3.2b.2c.2a=24abc\)

\(\Rightarrow A=24\) (Vì đề ra \(abc=1\))

Ta có :

\(\left(ax+b\right)\left(x^2-2cx+abc\right)=x^3-4x^2+3x+\frac{9}{5}\)

\(\Leftrightarrow ax^3+2acx^2+bx^2-2bcx+ab^2c=x^3-4x^2+3x+\frac{9}{5}\)

\(\Leftrightarrow ax^3+\left(2ac+b^2\right)x^2+\left(a^2bc-2bc\right)x+ab^2c=x^3-4x^2+3x+\frac{9}{5}\)

Đồng nhất hệ số ta được :

a = 1

2ac + b² = -4

a²bc - 2bc = 3

\(ab^2c=\frac{9}{5}\)

\(\Rightarrow a=1;b=\frac{3}{5};c=5\)

Tl

bạn  T I k cho tui trước tui trả lời cho

#Kirito

mệnh đè 2 sai

5(x²-9)+(2x+3)²=5x²-45+4x²+12x+9=9x²+12x-36=3(3x²+4x-12)

\(C=\left(x^2-2xy+2y^2\right)\left(x^2+y^2\right)-2x^3y-3x^3y^2+2xy^3\)

\(=x^4+3x^2y^2-2x^3y-2x^3y+y^4+2x^3y-3x^2y^2+2xy^3\)

\(=x^4+y^4\)

Bài 8:

a) \(x^2-2x+1=25\)

\(\Rightarrow\left(x-1\right)^2=25\)

\(\Rightarrow\orbr{\begin{cases}\left(x-1\right)^2=5^2\\\left(x-1\right)^2=\left(-5\right)^2\end{cases}}\Rightarrow\orbr{\begin{cases}x-1=5\\x-1=-5\end{cases}}\Rightarrow\orbr{\begin{cases}x=6\\x=-4\end{cases}}\)

b) \(\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\)

\(\Rightarrow25x^2+10x+1-25x^2+9=30\)

\(\Rightarrow10x+10=30\)

\(\Rightarrow10x=20\)

\(\Rightarrow x=2\)

c) \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+2\right)\left(x-2\right)=5\)

\(\Rightarrow x^3-1-x\left(x^2-4\right)=5\)

\(\Rightarrow x^3-1-x^3+4x=5\)

\(\Rightarrow4x=6\)

\(\Rightarrow x=\frac{3}{2}\)

d) \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=15\)

\(\Rightarrow x^3-6x^2+12x-8-\left(x^3-27\right)+6\left(x^2+2x+1\right)=15\)

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

\(\Rightarrow24x+25=15\)

\(\Rightarrow24x=-10\)

\(\Rightarrow x=\frac{-5}{12}\)

Bài 9:

a) \(-x^2+6x-15=-x^2+6x-9-6=-\left(x^2-6x+9\right)-6=-\left(x-3\right)^2-6\)

Ta có: \(\left(x-3\right)^2\ge0\Rightarrow-\left(x-3\right)^2\le0\Rightarrow-\left(x-3\right)^2-6\le-6\)

\(\Rightarrow-x^2+6x-15\) luôn âm với mọi \(x\)

b) \(-9x^2+24x-18=-9x^2+24x-16-2=-\left(9x^2-24x+16\right)-2=-\left(3x-4\right)^2-2\)

Ta có: \(\left(3x-4\right)^2\ge0\Rightarrow-\left(3x-4\right)^2\le0\Rightarrow-\left(3x-4\right)^2-2\le2\)

\(\Rightarrow-9x^2+24x-18\) luôn âm với mọi \(x\)

c) \(\left(x-3\right)\left(1-x\right)-2=x\left(1-x\right)-3\left(1-x\right)-2=x-x^2-3+3x-2=-x^2+4x-5\)

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

Ta có: \(\left(x-1\right)^2\ge0\Rightarrow-\left(x-1\right)^2\le0\Rightarrow-\left(x-1\right)^2-1\le-1\)

\(\Rightarrow\left(x-3\right)\left(1-x\right)-2\) luôn âm với mọi x

d) \(\left(x+4\right)\left(2-x\right)-10=x\left(2-x\right)+4\left(2-x\right)-10=2x-x^2+8-4x-10=-x^2-2x-2\)

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

Ta có: \(\left(x+1\right)^2\ge0\Rightarrow-\left(x+1\right)^2\le0\Rightarrow-\left(x+1\right)^2-1\le-1\)

\(\Rightarrow\left(x+4\right)\left(2-x\right)-10\) luôn âm với mọi \(x\)