1 What an interesting film it is!

2 How fine the weather is!

3 My bathroom has a sink, a tub and a showers

4 Mrs Na grows vegetables and raises cattles on her farm

5 Which is the most suitable apartment for your family

1.What an interesting film it is!

2.How fine the weather is!

3.My bathroom has a sink, a tub and a showers.

4.Mrs Na grows vegetables and raises cattles on her farm.

5.Which is the most suitable apartment for your family.

e: Ta có: \(\left(x+1\right)\left(x+2\right)=444222\)

\(\Leftrightarrow x^2+3x-444220=0\)

\(\text{Δ}=3^2-4\cdot1\cdot\left(-444220\right)=1776889\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là 

\(\left\{{}\begin{matrix}x_1=\dfrac{-3-1333}{2}=-668\\x_2=\dfrac{-3+1333}{2}=665\end{matrix}\right.\)

lớp 6 be like ;-;

Nhìu thiệt á, tách bớt ra đi nhé!

1 D

2 D

3 D

4 D

5 A

1. D

2. D

3. D

4. D

5.A 

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

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

Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)

\(A=\left(x^2-6x+9\right)-7=\left(x-3\right)^2-7\ge7\\ A_{min}=7\Leftrightarrow x=3\\ B=\left(9x^2+6x+1\right)-4=\left(3x+1\right)^2-4\ge-4\\ B_{min}=-4\Leftrightarrow x=-\dfrac{1}{3}\\ C=\left(x^2-2\cdot\dfrac{5}{2}x+\dfrac{25}{4}\right)-\dfrac{9}{4}=\left(x-\dfrac{5}{2}\right)^2-\dfrac{9}{4}\ge-\dfrac{9}{4}\\ C_{min}=-\dfrac{9}{4}\Leftrightarrow x=\dfrac{5}{2}\\ D=\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\\ D_{min}=\dfrac{3}{4}\Leftrightarrow x=\dfrac{1}{2}\)

\(E=3\left(x^2+2\cdot\dfrac{1}{3}x+\dfrac{1}{9}\right)-\dfrac{4}{3}=3\left(x+\dfrac{1}{3}\right)^2-\dfrac{4}{3}\ge-\dfrac{4}{3}\\ E_{min}=-\dfrac{4}{3}\Leftrightarrow x=-\dfrac{1}{3}\\ F=x^2-2x+1+x^2-4x+4+2021\\ F=2\left(x^2-3x+\dfrac{9}{4}\right)+\dfrac{4031}{2}=2\left(x-\dfrac{3}{2}\right)^2+\dfrac{4031}{2}\ge\dfrac{4031}{2}\\ F_{min}=\dfrac{4031}{2}\Leftrightarrow x=\dfrac{3}{2}\)

a) Ta có: MN⊥d, EF⊥d

=> MN//EF(từ vuông góc đến song song)

b) Ta có: \(\widehat{MPQ}=180^0-\widehat{MPb}=180^0-55^0=125^0\)(kề bù)

\(\Rightarrow\widehat{MPQ}=\widehat{NMc}=125^0\)

Mà 2 góc này đồng vị

=> PQ//MN

Mà MN//EF

=> PQ//EF

Cảm ơn bạn. Bạn có thể giúp mình làm nốt câu c được không?

T T F T T F