giúp mik vs ạ :((( nhiều bài quá mik làm hỏng kịp
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![](https://rs.olm.vn/images/avt/0.png?1311)
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4.2:
a: x^2-x+1=x^2-x+1/4+3/4
=(x-1/2)^2+3/4>=3/4>0 với mọi x
=>x^2-x+1 ko có nghiệm
b: 3x-x^2-4
=-(x^2-3x+4)
=-(x^2-3x+9/4+7/4)
=-(x-3/2)^2-7/4<=-7/4<0 với mọi x
=>3x-x^2-4 ko có nghiệm
5:
a: x^2+y^2=25
x^2-y^2=7
=>x^2=(25+7)/2=16 và y^2=16-7=9
x^4+y^4=(x^2)^2+(y^2)^2
=16^2+9^2
=256+81
=337
b: x^2+y^2=(x+y)^2-2xy
=1^2-2*(-6)
=1+12=13
x^3+y^3=(x+y)^3-3xy(x+y)
=1^3-3*1*(-6)
=1+18=19
![](https://rs.olm.vn/images/avt/0.png?1311)
1 interested in travelling by plane
2 15 minutes riding his bike to school
3 is made by my mother everymorning
4 take care of her little brother
5 able to speak 2 languages when he was young
6 able to play the piano
7 is played all over VN
8 to play voleyball very well when he was young
9 to go fishing when he was young
10 way to the shopping mall
![](https://rs.olm.vn/images/avt/0.png?1311)
over
camping
was
couldn't
percussion
takes
look after
go around
go through
wind
how long
for
do
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài cuối mình không thấy rõ đề nhưng mình đoán là thế này bạn nhé.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 5:
a: \(=\dfrac{a+2\sqrt{a}+a-2\sqrt{a}}{a-4}\cdot\dfrac{a-4}{2\sqrt{a}}=\dfrac{2a}{2\sqrt{a}}=\sqrt{a}\)
b: Để A-2>0 thì căn a-2>0
=>căn a>2
=>a>4
c: Để 4/A+1 là số nguyên thì \(\sqrt{a}+1\inƯ\left(4\right)\)
=>\(\sqrt{a}+1\in\left\{1;2;4\right\}\)
=>\(a\in\left\{1;9\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1 does this red dress cost
2 is this shirt
does this short cost
3 do these shoes cost
4 has brown backpack
5 how to get to the bus stop
6 to stay in bed
7 are cleaned by Nga everyday
8 is played with a bow
9 worked for that company for 10 years
10 15 minutes walking to school
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left|2x-3\right|=3-2x\)
\(ĐK:x\le\dfrac{3}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=3-2x\\3-2x=3-2x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\0=0\left(đúng\right)\end{matrix}\right.\)
Vậy \(S=\left\{x\in R;x=\dfrac{3}{2}\right\}\)
Câu 18: A
Câu 19: B