a: =(x^2+3x)(x^2+3x+2)+1

=(x^2+3x)^2+2(x^2+3x)+1

=(x^2+3x+1)^2>=0 với mọi x

b: (a^2+b^2+c^2)(x^2+y^2+z^2)-(ax+by+cz)^2

=a^2x^2+a^2y^2+a^2z^2+b^2x^2+b^2y^2+b^2z^2+c^2x^2+c^2y^2+c^2z^2-a^2x^2-b^2y^2-c^2z^2-2axby-2axcz-2bycz

=(a^2y^2-2axby+b^2x^2)+(a^2z^2-2azcx+c^2x^2)+(b^2z^2-2bzcy+c^2y^2)

=(ay-bx)^2+(az-cx)^2+(bz-cy)^2>=0(luôn đúng)

bài 1:

a, x + |2 - x| = 6

=> |2 - x| = 6 - x (1)

=>\(\orbr{\begin{cases}2-x=6-x\\2-x=x-6\end{cases}}\Rightarrow\orbr{\begin{cases}2=6\left(ktm\right)\\x=4\left(tm\right)\end{cases}}\)

b. |x - 7| = 7 

=> \(\orbr{\begin{cases}x-7=7\\x-7=-7\end{cases}\Rightarrow\orbr{\begin{cases}x=14\left(ktm\right)\\x=0\left(tm\right)\end{cases}}}\)

c, Tương tự b

bài 2:

a, Vì \(\hept{\begin{cases}\left|x+2\right|\ge0\\\left|y+5\right|\ge0\end{cases}}\forall x,y\Rightarrow\left|x+2\right|+\left|y+5\right|\ge0\) (1)

Mà |x + 2| + |y + 5| = 0 (2)

Từ (1),(2) => \(\hept{\begin{cases}x+2=0\\y+5=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-2\\y=-5\end{cases}}\)

b, tương tự a

1)

a) x + | 2 - x | = 6

\(\Rightarrow\)| 2 - x | = 6 - x

\(\Rightarrow\)\(\orbr{\begin{cases}2-x=6-x\\2-x=x-6\end{cases}}\)\(\Rightarrow\)\(\orbr{\begin{cases}2=6\\x=4\end{cases}}\)

b) | x - 7 | = 7

x - 7 = +;- 7

\(\Rightarrow\)\(\orbr{\begin{cases}x-7=7\\x-7=-7\end{cases}}\)\(\Rightarrow\)\(\orbr{\begin{cases}x=14\\x=0\end{cases}}\)

c) | x + 1 | = 5

x + 1 = +;- 5

\(\Rightarrow\)\(\orbr{\begin{cases}x+1=5\\x+1=-5\end{cases}}\)\(\Rightarrow\)\(\orbr{\begin{cases}x=4\\x=-6\end{cases}}\)

2) Tự làm :v

khong dua nao tra loi a . tra loi di chi cho 5 h

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

\(\left\{{}\begin{matrix}a=1\\b=1\\x=\sqrt{x}\\y=\sqrt{y}\end{matrix}\right.\)

áp vào \(\left(1.\sqrt{x}+1.\sqrt{2-x}\right)^2\le\left(1^2+1^2\right)\left(\sqrt{x}^2+\sqrt{2-x}^2\right)=2.\left(x+2-x\right)=2.2=4\)\(\left(1.\sqrt{x}+1.\sqrt{2-x}\right)^2\le4\Rightarrow\left(1.\sqrt{x}+1.\sqrt{2-x}\right)\le2\)

tại đâu bạn tự tìm cho vui

Theo bài ra ta có:

\(\left(a^2+b^2\right)\left(x^2+y^2\right)\ge\left(ax+by\right)^2\)

\(\Leftrightarrow a^2x^2+a^2y^2+b^2x^2+b^2y^2\ge a^2x^2+2axby+b^2y^2\)

\(\Leftrightarrow a^2y^2-2axby+b^2x^2\ge0\)

\(\Leftrightarrow\left(ay-bx\right)^2\ge0\left(true\right)\)

a) \(5x^2-4x=9\)

\(5x^2-4x-9=0\)

\(5x^2+5x-9x-9=0\)

\(5x\left(x+1\right)-9\left(x+1\right)=0\)

\(\left(x+1\right)\left(5x-9\right)=0\)

\(\hept{\begin{cases}x+1=0\\5x-9=0\end{cases}}\)

\(\hept{\begin{cases}x=-1\\x=\frac{9}{5}\end{cases}}\)

b) \(4x^2-2x+\frac{1}{4}\) với x = 0,25

Thay x = 0,25 vào biểu thức, ta có:

\(4.\left(0,25\right)^2-2.\left(0,25\right)+\frac{1}{4}=0\)