1 You should not wear shorts when going to the pagoda

2 At Tet, our house is more beautifully decorated than during the year

3 Sitting in front of a computer all day can cause health problems

4 Snow White is very kind to people and animals

5 Hung King Temple festival has been a public holiday in VN since 2007

III

1 Last night, we were having dinner when the telephone rang

2 Life in the countryside has changed a lot over the past ten years

3 Nam doesn't mind listening to classical music

Bày đặt sad mặt tâm trạng đồ😂

làm màu các thứ =)))

\(\Leftrightarrow\left\{{}\begin{matrix}x-2y+z=2\\4y-4z=-6\\-y+z=-13\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-2y+z=2\\y-z=-\dfrac{3}{2}\\-y+z=-13\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-2y+z=2\\y-z=-\dfrac{3}{2}\\0=-\dfrac{29}{2}\end{matrix}\right.\)

Hệ đã cho vô nghiệm

có thể chỉ rõ cho chút ko ạk

Bài 1.

a. $=a^2+2.a.12+12^2=a^2+24a+144$

b. $=(3a)^2+2.3a.\frac{1}{3}+(\frac{1}{3})^2=9a^2+2a+\frac{1}{9}$

c. $=(5a^2)^2+2.5a^2.6+6^2=25a^4+60a^2+36$

d. $=\frac{1}{4}+2.\frac{1}{2}.4b+(4b)^2$

$=\frac{1}{4}+4b+16b^2$

e.

$=(a^m)^2+2.a^m.b^n+(b^n)^2$

$=a^{2m}+2a^mb^n+b^{2n}$

Bài 2.

$(x-0,3)^2=x^2-0,6x+0,09$

$(6x-3y)^2=36x^2-36xy+9y^2$

$(5-2xy)^2=25-20xy+4x^2y^2$
$(x^4-1)^2=x^8-2x^4+1$

$(x^5-y^3)^2=x^{10}-2x^5y^3+y^6$

1: \(\begin{cases}x-2y+z=2\\ x+2y-3z=-4\\ x-3y+2z=-11\end{cases}\)

=>\(\begin{cases}x-2y+z-x-2y+3z=2+4=6\\ x-2y+z-x+3y-2z=2-\left(-11\right)=2+11=13\\ x-2y+z=2\end{cases}\)

=>\(\begin{cases}-4y+4z=6\\ y-z=13\\ x-2y+z=2\end{cases}\Rightarrow\begin{cases}y-z=-1,5\\ y-z=13\\ x-2y+z=2\end{cases}\)

=>(x;y;z)∈∅

Câu 2: Để hệ có nghiệm duy nhất thì \(\frac{2}{m}<>\frac{m-1}{3}\)

=>\(m\left(m-1\right)<>6\)

=>\(m^2-m-6<>0\)

=>(m-3)(m+2)<>0

=>m∉{3;-2}

\(\begin{cases}2x+\left(m-1\right)y=4\\ mx+3y=m+3\end{cases}\Rightarrow\begin{cases}2mx+\left(m^2-m\right)y=4m\\ 2mx+6y=2m+6\end{cases}\)

=>\(\begin{cases}2mx+\left(m^2-m\right)y-2mx-6y=4m-2m-6\\ mx+3y=m+3\end{cases}\)

=>\(\begin{cases}y\left(m^2-m-6\right)=2m-6\\ mx+3y=m+3\end{cases}\Rightarrow\begin{cases}y=\frac{2\left(m-3\right)}{\left(m-3\right)\left(m+2\right)}=\frac{2}{m+2}\\ mx=m+3-3y=m+3-\frac{6}{m+2}=\frac{m^2+5m+6-6}{m+2}=\frac{m^2+5m}{m+2}\end{cases}\)

=>\(\begin{cases}y=\frac{2}{m+2}\\ x=\frac{m+5}{m+2}\end{cases}\)

x-3y=2

=>\(\frac{m+5}{m+2}-\frac{6}{m+2}=2\)

=>\(\frac{m-1}{m+2}=2\)

=>2(m+2)=m-1

=>2m+4=m-1

=>m=-5(nhận)

Bài 13:

$6-2\sqrt{5}=5-2\sqrt{5}.\sqrt{1}+1$

$=(\sqrt{5}-1)^2$

Tương tự: $6+2\sqrt{5}=(\sqrt{5}+1)^2$
Do đó:
$M=\sqrt{(\sqrt{5}+1)^2}-\sqrt{(\sqrt{5}-1)^2}$

$=|\sqrt{5}+1|-|\sqrt{5}-1|=(\sqrt{5}+1)-(\sqrt{5}-1)$

$=2$

Bài 14:

a.

$M=\sqrt{4+2\sqrt{4}.\sqrt{5}+5}-\sqrt{4-2\sqrt{4}.\sqrt{5}+5}$

$=\sqrt{(\sqrt{4}+\sqrt{5})^2}-\sqrt{(\sqrt{4}-\sqrt{5})^2}$

$=|\sqrt{4}+\sqrt{5}|-|\sqrt{4}-\sqrt{5}|$

$=2+\sqrt{5}-(\sqrt{5}-2)=4$

b.

$N=\sqrt{7-2\sqrt{7}+1}-\sqrt{7+2\sqrt{7}+1}$

$=\sqrt{(\sqrt{7}-1)^2}-\sqrt{(\sqrt{7}+1)^2}$

$=|\sqrt{7}-1|-|\sqrt{7}+1|$

$=(\sqrt{7}-1)-(\sqrt{7}+1)=-2$