Câu 1

a)$(2x-1)^2=4x^2-4x+1$

b)$\bigg(x+\dfrac{1}{2}\bigg)^2=x^2+x+\dfrac{1}{4}$

c)$(2x+3)^2=4x^2+12x+9$

d)$(x-4)^2=x^2-8x+16$

e)$(2x-3y)(4x+6y)=(2x-3y)(2x+3y).2\\ =2(4x^2-9y^2)=8x^2-18y^2$

f)$2023^2-2021^2=(2023-2021)(2023+2021)\\ =2.4044=8088$

g)$105^2-2.525+25=105^2-2.105.5+5^2\\ =(105-5)^2=100^2=10000$

h)$97.103=(100-3)(100+3)\\ =100^2-3^2=10000-9=9991$

Câu 2

a)$25x^2-30x+9\\=(5x)^2-2.5x.3+3^2\\=(5x-3)^2\\=(5.2-3)^2\\=7^2=49$

b)$4x^2-28x+49\\=(2x)^2-2.2x.7+7^2\\=(2x-7)^2\\=(2.4-7)^2=1^2=1$

Câu 3

a)$(5-2x)^2-4^2=0\\ \Leftrightarrow (5-2x-4)(5-2x+4)=0\\ \Leftrightarrow \left[\begin{matrix}1-2x=0\\9-2x=0\end{matrix}\right.\\ \Leftrightarrow \left[\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{9}{2}\end{matrix}\right.$

b)$x^2-2x+1=25\\ \Leftrightarrow (x-1)^2=25\\ \Leftrightarrow \left[\begin{matrix}x-1=-5\\x-1=5\end{matrix}\right.\\ \Leftrightarrow \left[\begin{matrix}x=-4\\x=6\end{matrix}\right.$

c)$3(x-1)^2-3x(x-5)=1\\ \Leftrightarrow 3x^2-6x+3-3x^2+15x=1\\ \Leftrightarrow 9x=-2\\ \Leftrightarrow x=\dfrac{-2}{9}$

Câu 4

a)$4x^2-20x+2050=(2x)^2-2.2x.5+5^2+2025\\=(2x-5)^2+2025$

Vì $(2x-5)\ge0\ \forall x\in R$ nên

$Min=2025$ khi $(2x-5)^2=0\Leftrightarrow x=\dfrac{5}{2}$

$9x^2+6x-2020=(3x)^2+2.3x.1+1^2-2021\\ =(3x+1)^2-2021$

Vì $(3x+1)^2\ge0\ \forall x\in R$ nên

$Min=-2021$ khi $(3x+1)^2=0\Leftrightarrow x=\dfrac{-1}{3}$

b)$12x-4x^2-11=-(4x^2-12x+9)-2\\ =-(2x-3)^2-2$

Vì $(2x-3)^2\ge0\ \forall x\in R$ nên $-(2x-3)^2\le0\ \forall x\in R$

$\Rightarrow Max=-2$ khi $(2x-3)^2=0\Leftrightarrow x=\dfrac{3}{2}$

$6x-9x^2+2025=-(9x^2-6x+1)+2026\\ =2026-(3x-1)^2$

Vì $(3x-1)^2\ge0\ \forall x\in R$ nên

$Max=2026$ khi $(3x-1)^2=0\Leftrightarrow x=\dfrac{1}{3}$

Xin 5 sao + ctlhn ạ

Đáp án:

Giải thích các bước giải:

Bài 1:

1) $(2x-1)^{2}=4x^{2}-4x+1$

2) $(x+\frac{1}{2})^{2}=x^{2}+x+\frac{1}{4}$

3) $(2x+3)^{2}=4x^{2}+12x+9$

4) $(x-4)^{2}=x^{2}-8x+16$

5) $(2x-3y)(4x+6y)=2(4x^{2}-9y^{2})=8x^{2}-18y^{2}$

6) $2023^{2}-2021^{2}=(2023-2021)(2023+2021)=2.4044=8088$

7) $105^{2}-2.525+25=(105-5)^{2}=100^{2}=10000$

8) $97.103=(100-3)(100+3)=100^{2}-3^{2}=10000-9=9991$

Bài 2:

a. $25x^{2}-30x+9=(5x-3)^{2}=(5.2-3)^{2}=7^{2}=49$

b. $4x^{2}-28x+49=(2x-7)^{2}=(2.4-7)^{2}=1^{2}=1$

Bài 4:

a. $4x^{2}-20x+2050=4x^{2}-20x+25+2025=(2x-5)^{2}+2025$

Mà $(2x-5)^{2}\geq0⇔(2x-5)^{2}+2025\geq2025$

Dấu "=" xảy ra khi $2x-5=0⇔x=\frac{5}{2}$

MIN = 2025 khi x=$\frac{5}{2}$

Tương tự ta cx có: MIN = -2021 khi x=-$\frac{1}{3}$

b. $12x-4x^{2}-11$

$-4x^{2}+12x-9-2=-(2x-3)^{2}-2$

Mà $-(2x-3)^{2}\leq0⇔-(2x-3)^{2}-2\leq-2$

Dấu "=" xảy ra khi $2x-3=0⇔x=\frac{3}{2}$

MAX = -2 khi x=$\frac{3}{2}$

Tương tự ta cx có: MAX = 2026 khi x=$\frac{1}{3}$