a/ \(x^2-6x+10=x^2-2.x.3+3^2+1=\left(x-3\right)^2+1\)

Với mọi x ta có :

\(\left(x-3\right)^2\ge0\)

\(\Leftrightarrow\left(x-3\right)^2+1>0\)

\(\Leftrightarrow x^2-6x+10>0\)

b/ \(x^2-4x+7=x^2-2.x.2+2^2+3=\left(x-2\right)^2+3\)

Với mọi x ta có :

\(\left(x-2\right)^2\ge0\)

\(\Leftrightarrow\left(x-2\right)^2+3\ge3\)

\(\Leftrightarrow x^2-4x+7\ge3\left(đpcm\right)\)

c/ \(x^2+x+1=x^2+2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\dfrac{3}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)

Với mọi x ta có :

\(\left(x+\dfrac{1}{2}\right)^2\ge0\)

\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)

\(\Leftrightarrow x^2+x+1>0\left(đpcm\right)\)

d/ \(x^2+y^2+4x-6y+15=\left(x^2+4x+2^2\right)+\left(y^2-6y+3^2\right)+2=\left(x+2\right)^2+\left(y-3\right)^2+2\)

Với mọi x,y ta có :

\(\left\{{}\begin{matrix}\left(x+2\right)^2\ge0\\\left(y-3\right)^2\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left(x+2\right)^2+\left(y-3\right)^2\ge0\)

\(\Leftrightarrow\left(x+2\right)^2+\left(y-3\right)^2+2\ge0\)

\(\Leftrightarrow x^2+y^2+4x-6y+15>0\left(đpcm\right)\)

2/ Ta có :

\(\left(a+b\right)^2-4ab=a^2+2ab+b^2-4ab=a^2-2ab+b^2=\left(a-b\right)^2\)

Vậy \(\left(a-b\right)^2=\left(a+b\right)^2-4ab\left(đpcm\right)\)

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

Mà \(x+y=7;xy=-3\)

\(\Leftrightarrow x^2+y^2=7^2-2.\left(-3\right)=49+6=55\)

2.

Ta có hằng đẳng thức : \(\left(a-b\right)^2=a^2-2ab+b^2\left(1\right)\)

Lại có  \(\left(a+b\right)^2=a^2+2ab+b^2\)

\(\Rightarrow\left(a+b\right)^2-4ab=a^2+2ab-4ab+b^2\)

\(\Leftrightarrow\left(a+b\right)^2-4ab=a^2-2ab+b^2\left(2\right)\)

Từ (1) và (2)  \(\Rightarrow\left(a-b\right)^2=\left(a+b\right)^2-4ab\)( đpcm )

3.

Ta có hằng đẳng thức  \(\left(x+y\right)^2=x^2+2xy+y^2\)

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

Thay  \(x+y=7\)và  \(xy=-3\)vào ta được :

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

\(\Leftrightarrow x^2+y^2=49+6=55\)

Vậy ...

1. 

a) Đặt  \(A=x^2-6x+10\)

\(A=\left(x^2-6x+9\right)+1\)

\(A=\left(x-3\right)^2+1\)

Mà  \(\left(x-3\right)^2\ge0\forall x\)

\(\Rightarrow A\ge1>0\)

Vậy ...

b) Đặt \(B=x^2-4x+7\)

\(B=\left(x^2-4x+4\right)+3\)

\(B=\left(x-2\right)^2+3\)

Mà  \(\left(x-2\right)^2\ge0\forall x\)

\(\Rightarrow B\ge3\)

Vậy ...

1.

c) Đặt \(C=x^2+x+1\)

\(C=\left(x^2+x+\frac{1}{4}\right)+\frac{3}{4}\)

\(C=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)

Mà  \(\left(x+\frac{1}{2}\right)^2\ge0\forall x\)

\(\Rightarrow C\ge\frac{3}{4}>0\)

Vậy ...

d) Đặt \(D=x^2+y^2+4x-6y+15\)

\(D=\left(x^2+4x+4\right)+\left(y^2-6y+9\right)\)

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

Mà  \(\left(x+2\right)^2\ge0\forall x\)

      \(\left(y-3\right)^2\ge0\forall y\)

\(\Rightarrow D\ge0\)

Dấu "=" xảy ra khi :  \(\hept{\begin{cases}x+2=0\\y-3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-2\\y=3\end{cases}}\)

\(A=x^2+y^2-4x+2y+7\)

\(=x^2+y^2-4x+2y+4+1+3\)

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

\(=\left(x-2\right)^2+\left(y+1\right)^2+3\)

Ta thấy: \(\left\{{}\begin{matrix}\left(x-2\right)^2\ge0\forall x\\\left(y+1\right)^2\ge0\forall y\end{matrix}\right.\)

\(\Rightarrow\left(x-2\right)^2+\left(y+1\right)^2\ge0\forall x,y\)

\(\Rightarrow\left(x-2\right)^2+\left(y+1\right)^2+3\ge3>0\forall x,y\)

A= x²+y²-4x+2y+7

= (x²-4x+4)+(y²+2y+1)+2

= (x-2)²+(y+1)²+2

Ta thấy: (x-2)²\(\ge0\)

(y+1)²\(\ge0\)

\(\Rightarrow\)(x-2)²+(y+1)²+2\(\ge2\)

\(\Rightarrow\)A\(\ge2\)

Vậy A>0 \(\forall x,y\)

\(A=x^2+y^2-4x+2y+7\)

\(=x^2+y^2-4x+2y+4+1+2\)

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

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

Ta thấy: \(\left\{{}\begin{matrix}\left(x-2\right)^2\ge0\forall x\\\left(y+1\right)^2\ge0\forall y\end{matrix}\right.\)

\(\Rightarrow\left(x-2\right)^2+\left(y+1\right)^2\ge0\forall x,y\)

\(\Rightarrow\left(x-2\right)^2+\left(y+1\right)^2+2\ge2>0\forall x,y\)

aを見つける= 175度はどれくらい尋ねるaを見つける= 175度はどれくらい尋ねる

b. \(4x^2+4x-7=0\)

\(\Leftrightarrow\left(2x\right)^2+2.2x+1-8=0\)

\(\Leftrightarrow\left(2x+1\right)^2-8=0\)

\(\Leftrightarrow\left(2x+1-\sqrt{8}\right)\left(2x+1+\sqrt{8}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x+1-\sqrt{8}=0\\2x+1+\sqrt{8}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{\sqrt{8}-1}{2}\\x=\frac{-\sqrt{8}-1}{2}\end{cases}}\)

k mình nha bn <3 thanks nhìu hen

a, Với f(0)=4.0³+0=0

    Với f(-0,5)=4.(-0.5)³+(-0,5)=-1

thế còn phần b

a,x^2-7x=0

<=>x(x-7)=0

<=>th1 x=0

th2 x-7=0=>x=7

vậy x=0 hoặc 7

\(a^2-7a=0\)

\(\Rightarrow a\left(a-7\right)=0\)

\(\Rightarrow\hept{\begin{cases}a=0\\a-7=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}a=0\\a=7\end{cases}}\)

b,2x-4x^2=0

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

<=> th1 2x=0=>x=0

th2 1-2x=0=>x=1/2

vậy x=0 hoặc 1/2

a) Ta có:

\(x^2-x+1\)

\(=x^2-2\cdot\dfrac{1}{2}\cdot x+\dfrac{1}{4}+\dfrac{3}{4}\)

\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)

Mà: \(\left(x-\dfrac{1}{2}\right)^2\ge0\) và \(\dfrac{3}{4}>0\) nên

\(\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\forall x\)

\(\Rightarrow x^2-x+1>0\forall x\)