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
