Question

1. Tìm x GPT:

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

2. Chứng minh biểu thức sau dương mọi a,b

\(2a^2+2ab+1+a+b^2\)

Bài tuy dễ nhá ( Bác nào làm đúng cho 2 GP )

Answer 1

1.

\(\left(x-1\right)^3-x\left(x-2\right)^2-x+2=0\\ x^3-3x^2+3x-1-x\left(x^2-4x+4\right)-x+2=0\\ x^3-3x^2+3x-1-x^3+4x^2-4x-x+2=0\\ x^2-2x+1=0\\ \Rightarrow\left(x-1\right)^2=0\\ \Rightarrow x-1=0\\ \Rightarrow x=1\)

Answer 2

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

\(\Leftrightarrow\left(x^3-3x^2+3x-1\right)-x\left(x^2-4x+4\right)-x+2=0\)

\(\Leftrightarrow x^3-3x^2+3x-1-x^3+4x^2-4x-x+2=0\)

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

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

Answer 3

2.

\(2a^2+2ab+1+a+b^2 \)

\(=\left(a^2+2ab+b^2\right)+\left(a^2+a+1\right)\)

\(=\left(a+b\right)^2+\left(a^2+2.a.\dfrac{1}{2}+\dfrac{1}{4}\right)+\dfrac{3}{4}\)

\(=\left(a+b\right)^2+\left(a+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)

+ Ta có :

\(\left(a+b\right)^2\ge0 ; \left(a+\dfrac{1}{2}\right)^2\ge0\)

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

\(\Rightarrow\left(a+b\right)^2+\left(a+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

\(\Rightarrow\left(a+b\right)^2+\left(a+\dfrac{1.}{2}\right)^2+\dfrac{3}{4}>0\)

Hay \(2a^2+2ab+1+a+b^2>0\forall a,b\)

Answer 4

1.

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

\(\Rightarrow\left(x-1\right)^3-x\left(x-2\right)^2-\left(x-2\right)=0\)

\(\Rightarrow\left(x-1\right)^3-\left(x\left(x-2\right)+\left(x-2\right)\right)=0\)

\(\Rightarrow-\left(x+1\right)^3-\left(x-2\right)\left(x+1\right)\) = 0

\(\Rightarrow-\left[\left(x+1\right)^3+\left(x-2\right)\left(x+1\right)\right]=0\)

\(\Rightarrow-\left(x+1\right)\left[\left(x+1\right)^2+\left(x-2\right)\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}-\left(x+1\right)=0\\\left(x+1\right)^2+\left(x-2\right)=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-1\\\left(-1+1\right)^2+\left(x-2\right)=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)

Answer 5

1. \(D=\left(x-1\right)^3-x\left(x-2\right)^2-x+2\)
\(D=\left(x-1\right)^3-x\left(x-2\right)^2-\left(x-2\right)\)
\(D=\left(x-1\right)^3-\left(x-2\right)\left[x\left(x-2\right)+1\right]\)
\(D=\left(x-1\right)^3-\left(x-2\right)\left[x^2-2x+1\right]\)
\(D=\left(x-1\right)^3-\left(x-2\right)\left(x-1\right)^2\)
\(D=\left(x-1\right)^2\left[\left(x-1\right)-\left(x-2\right)\right]\)
\(D=\left(x-1\right)^2\left[\left(x-1-x+2\right)\right]\)
\(D=\left(x-1\right)^2\left[1\right]\)
\(D=\left(x-1\right)^2\)
\(D=0\Rightarrow\left(x-1\right)^2=0\)
\(x=1\)

KL: x = 1

Answer 6

2.

C/m \(2a^2+2ab+1+a+b^2>0,\forall a,b\)

\(VT=a^2+2ab+b^2+a^2+a+1\)

\(=\left(a+b\right)^2+a^2+a+\dfrac{1}{4}+\dfrac{3}{4}\)

\(=\left(a+b\right)^2+\left(a+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\)

\(\Rightarrowđpcm\)

Answer 7

Ta có :

\(2a^2+2ab+1+a+b^2\)

\(=a^2+a^2+2ab+1+a+b\\ =\left(a^2+2ab+b^2\right)+\left(a^2+a+1\right)\\ =\left(a+b\right)^2+\left(a+\dfrac{1}{2}\right)^2+\dfrac{1}{4}\)

Với mọi \(x\in R\) , ta có : \(\left(a+b\right)^2\ge0;\left(a+\dfrac{1}{2}\right)^2\ge0\)

\(\Rightarrow\left(a+b\right)^2+\left(a+\dfrac{1}{2}\right)^2+\dfrac{1}{4}\ge\dfrac{1}{4}\)

Vậy biểu thức trên dương với mọi a,b

Answer 8

Ta có :

\(2a^2+2ab+1+a+b^2\)

\(=a^2+a^2+2ab+1+a+b\)

\(=(a^2+2ab+b^2)+(a^2+a+1)\)

\(=(a+b)^2+(a+1/2)^2+3/4\)

Với mọi \(\text{ x∈R}\) , ta có : \((a+b)^2 ≥ 0;(a+1/2)^2 ≥0 \)

\(⇒(a+b)^2+(a+1/2)^2+3/4 ≥ 3/4\)

Vậy biểu thức trên luôn dương với mọi a,b

Answer 9

Sao nói là cho 2 GP mà thấy có người đc bn tik lại ko đc GP nào luôn vậy

Answer 10

Bài1:

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

=>\(x^3-3x^2+3x-1-x^2+2x-x+2=0\)

=>\(x^3-4x^2+4x+1=0\)

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

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

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

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

Với mọi x thì \(\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>=\dfrac{3}{4}>0\)

Do đó:\(x+1=0\)

=>\(x=-1\)

Vậy...