Giúp mk 2 bài này ạ
Giúp mình 1 trong 2 bài này với ạ. Nếu được thì giải 2 bài này giúp mk với. Mình đang cần gấp lắm ạ 😢
jimmmmmmmmmmmmmmmmmmmmmmmmmmm
he he he he he he
bài 1:
bn lấy giá trị của √(4^2-3,9^2) là dc
bài 2
AB+BC=2√(3^2+4^2)=??
giúp mk 2 bài này với ạ,mk đang cần gấp
Bài 18:
a: Ta có: \(P=\left(\dfrac{\sqrt{a}}{2}-\dfrac{1}{2\sqrt{a}}\right)^2\cdot\left(\dfrac{\sqrt{a}-1}{\sqrt{a}+1}-\dfrac{\sqrt{a}+1}{\sqrt{a}-1}\right)\)
\(=\dfrac{\left(\sqrt{a}-1\right)^2\cdot\left(\sqrt{a}+1\right)^2}{4a}\cdot\dfrac{a-2\sqrt{a}+1-a-2\sqrt{a}-1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{\left(a-1\right)\cdot\left(-4\right)\cdot\sqrt{a}}{4a}\)
\(=\dfrac{-a+1}{\sqrt{a}}\)
b: Để P<0 thì -a+1<0
\(\Leftrightarrow-a< -1\)
hay a>1
c: Để P=-2 thì \(-a+1=-2\sqrt{a}\)
\(\Leftrightarrow-a+1+2\sqrt{a}=0\)
\(\Leftrightarrow a-2\sqrt{a}+1=2\)
\(\Leftrightarrow\left(\sqrt{a}-1\right)^2=2\)
\(\Leftrightarrow\sqrt{a}-1=\sqrt{2}\)
hay \(a=3+2\sqrt{2}\)
Bài 17:
a: Ta có: \(P=\dfrac{a\sqrt{a}-1}{a-\sqrt{a}}-\dfrac{a\sqrt{a}+1}{a+\sqrt{a}}+\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right)\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}+\dfrac{\sqrt{a}-1}{\sqrt{a}+1}\right)\)
\(=\dfrac{a+\sqrt{a}+1-a+\sqrt{a}-1}{\sqrt{a}}+\dfrac{a-1}{\sqrt{a}}\cdot\dfrac{a+2\sqrt{a}+1+a-2\sqrt{a}+1}{a-1}\)
\(=2+\dfrac{2a+2}{\sqrt{a}}\)
\(=\dfrac{2a+2\sqrt{a}+2}{\sqrt{a}}\)
Giải giúp mk 2 bài này vs ạ ,lý giải vì sao giúp mk luôn nhà ,mk cảm ơn
Part 6 :
1. especially
2. with
3. goods
4. perform
Part 7:
1. T
2. F
3. F
4.T
cc giúp mk 2 bài này với ạ
Bài 13:
a: Ta có: \(P=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\dfrac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}\)
b: Để \(P=\dfrac{1}{2}\) thì \(-10\sqrt{x}+4=\sqrt{x}+3\)
\(\Leftrightarrow-11\sqrt{x}=-1\)
hay \(x=\dfrac{1}{121}\)
Giải giúp mk 2 bài này với ạ mk đag cần gấp . dịch giúp mk 4 câu của part 7
P6:
1. big
2. been
3. bought
4. people
P7:
1. T
2. F
3. F
4. F
Mn làm giúp mk 2 bài này với ạ
1 Because she went to bed late last night, she couldn't go to school on time
2 It rained heavily so we couldn't go to the theater
3 The wind was strong, so the tree in my garden was uprooted
4 He fell asleep while driving, so he had an accident
5 If I had a guitar here, I could sing you a song
6 If I didn't need to finish my assignment, I could help you with your homework
7 If you don't study hard, you may fail the exam
8 If there is no oil in the engine, the car will breakdown
9 If it hadn't been for your help, I couldn't have finished the report
10 If it hadn't been for her disapprova, he would have been given that job
11 The house seemed as if it had been occupied for years
Giúp mk vs ạ, mk ko bt lm bài này ạ
Vì \(AB//CD\) nên \(\left\{{}\begin{matrix}\widehat{B}+\widehat{C}=180^0\\\widehat{A}+\widehat{D}=180^0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\widehat{B}=\left(180^0+40^0\right):2=110^0\\3\widehat{D}=180^0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\widehat{C}=180^0-110^0=70^0\\\widehat{D}=60^0\end{matrix}\right.\Rightarrow\widehat{A}=120^0\)
\(\widehat{B}=110^0\)
\(\widehat{C}=70^0\)
\(\widehat{A}=120^0\)
\(\widehat{D}=60^0\)
Giúp mk bài này với ạ, mk đg cần gấp:
Tìm x:
(x + 4)(x+2) - x2 =7
\(\Leftrightarrow x^2+6x+8-x^2=7\\ \Leftrightarrow6x=-1\Leftrightarrow x=-\dfrac{1}{6}\)
(x + 4)(x+2) - x2 =7
x2+ 2x + 4x + 8 - x2 = 7
6x + 8 = 7
6x = 7 - 8 = -1
=> x = \(\dfrac{-1}{6}\)
Ai giải giúp mk bài này với ạ, mk đang cần gấp:
(x + 2)2 - 9 = 0
Mk cảm ơn trc:3
\(\left(x+2\right)^2-9=0\)
\(\Rightarrow\left(x+2\right)^2=9\)
\(\Rightarrow\left[{}\begin{matrix}x+2=3\\x+2=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)
Vậy...