Tìm GTNN của :  \(x^2-4x+3\)

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

Vì  \(\left(x-2\right)^2\ge0\)  nên  \(\left(x-2\right)^2-1\ge-1\)

Vậy GTNN của biểu thức là -1 . Dấu bằng xảy ra khi x = 2

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

\(=2x^2+10x-x-5-3.\left(x^2-4x+4\right)+x^2-16\)

\(=2x^2+9x-5-3x^2+12x-12+x^2-16=21x-33\)

Khi x = -2 thì A = 21 . (-2) -33 = -75

Cảm ơnb bạnnha

\(ĐKXĐ:\hept{\begin{cases}x\ne\pm1\\x\ne-\frac{1}{2}\end{cases}}\)

a) \(A=\left(\frac{1}{x-1}+\frac{x}{x^3-1}\cdot\frac{x^2+x+1}{x+1}\right):\frac{2x+1}{x^2+2x+1}\)

\(\Leftrightarrow A=\left(\frac{1}{x-1}+\frac{x}{\left(x-1\right)\left(x+1\right)}\right):\frac{2x+1}{\left(x+1\right)^2}\)

\(\Leftrightarrow A=\frac{x+1+x}{\left(x-1\right)\left(x+1\right)}\cdot\frac{\left(x+1\right)^2}{2x+1}\)

\(\Leftrightarrow A=\frac{\left(2x+1\right)\left(x+1\right)}{\left(x-1\right)\left(2x+1\right)}\)

\(\Leftrightarrow A=\frac{x+1}{x-1}\)

b) Thay \(x=\frac{1}{2}\)vào A, ta được :

\(A=\frac{\frac{1}{2}+1}{\frac{1}{2}-1}=\frac{\frac{3}{2}}{-\frac{1}{2}}=-3\)

\(A=2x+xy^2-x^2y-2y\)

\(=2\left(x-y\right)-xy\left(x-y\right)\)

\(=\left(x-y\right)\left(2-xy\right)\)

\(=\left(-\dfrac{1}{2}-\dfrac{-1}{3}\right)\left(2-\dfrac{-1}{2}\cdot\dfrac{-1}{3}\right)\)

\(=\left(\dfrac{1}{3}-\dfrac{1}{2}\right)\cdot\left(2-\dfrac{1}{6}\right)\)

\(=\dfrac{-1}{6}\cdot\dfrac{11}{6}=-\dfrac{11}{36}\)

a,\(A=x^2-3x+5=x^2-2.\dfrac{3}{2}x+\dfrac{9}{4}+\dfrac{11}{4}=\)

\(=\left(x-\dfrac{3}{2}\right)^2+\dfrac{11}{4}\)

Do \(\left(x-\dfrac{3}{2}\right)^2\ge0\left(\forall x\right)\Rightarrow\left(x-\dfrac{3}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\left(\forall x\right)\)

Daau "=" xảy ra \(\Leftrightarrow\left(x-\dfrac{3}{2}\right)^2=0\Leftrightarrow x=\dfrac{3}{2}\)

Vaay \(MinA=\dfrac{11}{4}\Leftrightarrow x=\dfrac{3}{2}\)

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

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

Do \(-\left(x-1\right)^2\le0\Rightarrow1-\left(x-1\right)^2\le1\left(\forall x\right)\)

Dau "=" xay ra \(\Leftrightarrow-\left(x-1\right)^2=0\Leftrightarrow x=1\)

Vay \(MaxA=1\Leftrightarrow x=1\)

B1:

\(a.301^2=\left(300+1\right)^2=300^2+2.300.1+1^2\\ =90000+600+1=90601\\ b.88^2+2.88.12+12^2=\left(88+12\right)^2=100^2=10000\\ c.99.100=100^2-100=10000-100=9900\\ d,153^2+94.153+47^2=153^2+2.153.47+47^2=\left(153+47\right)^2=200^2=40000\)

B2:

\(A=x^2-20x+101\\ =x^2-2.x.10+10^2+1\\ =\left(x-10\right)^2+1\ge1\forall x\in R\left(Vì:\left(x-10\right)^2\ge0\forall x\in R\right)\\ \Rightarrow min_A=1\Leftrightarrow x-10=0\Leftrightarrow x=10\)

a: Ta có: \(A=x^2+2x+5\)

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

\(=\left(x+1\right)^2+4\ge4\forall x\)

Dấu '=' xảy ra khi x=-1

Ta có: (2x+3)(5x-1) - 5x(2x-7) = (2x+3)(5x-1) - 5x( 2x+3-10) 

= (2x+3)(5x-1) - 5x(2x+3) + 50x

= (2x+3) (5x - 1 - 5x) + 50x

= (2x+3) .-1 +50x

Thay x =0 => (2.0+3) . -1 +50.0 = -3

Tách ra mỗi câu một lần.

Dài quá không ai làm đâu.

Nhìn nản lắm.

Câu 3: 

a: \(49^2=2401\)

b: \(51^2=2601\)

c: \(99\cdot100=9900\)