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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)=??

Bài 5: 

a: BC=10cm

b: HA=4,8cm

HB=3,6(cm)

HC=6,4(cm)

Bài 6:

\(x^3=6+3\sqrt[3]{\left(3+2\sqrt{2}\right)\left(3-2\sqrt{2}\right)}\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right)\\ \Leftrightarrow x^3=6+3x\sqrt[3]{1}\\ \Leftrightarrow x^3-3x=6\\ y^3=34+3\sqrt[3]{\left(17+12\sqrt{2}\right)\left(17-12\sqrt{2}\right)}\left(\sqrt[3]{17+12\sqrt{2}}+\sqrt[3]{17-12\sqrt{2}}\right)\\ \Leftrightarrow y^3=34+3y\sqrt[3]{1}\\ \Leftrightarrow y^3-3y=34\\ \Leftrightarrow P=x^3-3x+y^3-3y+1980=6+34+1980=2020\)

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I

1 will have

2 would have

3 have

4 had

5 will make

6 would make

7 has

8 had

9 is

10 were

II

1 have

2 had

3 had had

4 will go

5 would go

6 would have gone

7 is

8 were - would visit

9 had been - would have visited

10 wouldn't be - were

I.

1. will have

2. would have

3. have

4. had

5. will made

6. would made

7. has

8. had

9. is

10. were

II.

1. have

2. had

3. had had

4. will go

5. would go

6. would have gone

7. is

8. were - would visit

9. had been - would have visited

10. will not be - is

lỗi

lỗi

Lỗi rồi bạn ơi

\(\dfrac{4}{7}+\dfrac{2}{9}+\dfrac{1}{4}+\dfrac{3}{7}+\dfrac{7}{9}+\dfrac{75}{100}\\ =\dfrac{4}{7}+\dfrac{2}{9}+\dfrac{1}{4}+\dfrac{3}{7}+\dfrac{7}{9}+\dfrac{3}{4}\\ =\left(\dfrac{4}{7}+\dfrac{3}{7}\right)+\left(\dfrac{2}{9}+\dfrac{7}{9}\right)+\left(\dfrac{1}{4}+\dfrac{3}{4}\right)\\ =\dfrac{7}{7}+\dfrac{9}{9}+\dfrac{4}{4}\\ =1+1+1\\ =3\)

Ta có: \(\dfrac{7}{11}=\dfrac{7\times3}{11\times3}=\dfrac{21}{33};\dfrac{8}{11}=\dfrac{8\times3}{11\times3}=\dfrac{24}{33}\)

2 phân số giữa là: \(\dfrac{22}{33};\dfrac{23}{33}\)

\(=\left(\dfrac{4}{7}+\dfrac{3}{7}\right)+\left(\dfrac{2}{9}+\dfrac{7}{9}\right)+\left(\dfrac{1}{4}+\dfrac{3}{4}\right)=1+1+1=3\)

(\(\dfrac{4}{7}\)+\(\dfrac{3}{7}\))+(\(\dfrac{2}{9}\)+\(\dfrac{7}{9}\))+(\(\dfrac{1}{4}\)+\(\dfrac{75}{100}\))

=1+1+1

=3

2.

\(n_{OH^-}=0,3\left(mol\right)\)

\(n_{H_3PO_4}=0,25\left(mol\right)\)

\(\Rightarrow T=\dfrac{6}{5}\Rightarrow\) Tạo 2 muối \(NaH_2PO_4,Na_2HPO_4\)

\(\Rightarrow\) Sản phẩm \(NaH_2PO_4,Na_2HPO_4,H_2O\)

PTHH: \(3NaOH+2H_3PO_4\rightarrow NaH_2PO_4+Na_2HPO_4+3H_2O\)

3.

a, Bảo toàn e:

\(3n_{NO}=3n_{Al}\Rightarrow n_{NO}=n_{Al}=0,05\left(mol\right)\)

\(\Rightarrow V=1,12\left(l\right)\)

\(\Rightarrow n_{HNO_3}=4n_{NO}=0,2\left(mol\right)\)

\(\Rightarrow C_M=\dfrac{0,2}{0,2}=1M\)

a: Thay x=2 vào Q, ta được:

\(Q=\dfrac{2-1}{2}=\dfrac{1}{2}\)

Câu 2: 

a) Ta có: \(AN=BN=\dfrac{AB}{2}\)(N là trung điểm của AB)

\(AM=MC=\dfrac{AC}{2}\)(M là trung điểm của AC)

mà AB=AC(ΔABC cân tại A)

nên AN=BN=AM=MC

Xét ΔBNC và ΔCMB có 

BN=CM(cmt)

\(\widehat{NBC}=\widehat{MCB}\)(hai góc ở đáy của ΔABC cân tại A)

BC chung

Do đó: ΔBNC=ΔCMB(c-g-c)

Câu 2: 

b) Ta có: ΔBNC=ΔCMB(cmt)

nên \(\widehat{NCB}=\widehat{MBC}\)(hai góc tương ứng)

hay \(\widehat{OBC}=\widehat{OCB}\)
Xét ΔOBC có \(\widehat{OBC}=\widehat{OCB}\)(cmt)

nên ΔOBC cân tại O(Định lí đảo của tam giác cân)

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