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Bài 5: 

Xét ΔABC vuông tại A có 

\(\widehat{B}=30^0\)

nên \(AC=\dfrac{1}{2}BC\)

=>BC=12cm

Ta có: ΔABC vuông tại A

mà AM là đường trung tuyến ứng với cạnh huyền BC

nên AM=BC/2=6(cm)

giúp em thêm b6 nữa dc ko ạ

giúp mik điiiiiiiii mad 

4.

\(A=\left\{1;2;3\right\}\) ; \(B=\left\{0;1;2;3\right\}\) ; \(C=[0;+\infty)\) ; \(D=\left\{\dfrac{1}{2};3\right\}\)

\(\Rightarrow A\subset B\) ; \(A\subset C\); \(B\subset C\) ; \(D\subset C\)

5.

\(A=\left\{\dfrac{1}{2};2\right\}\) ; \(B=\left\{\dfrac{1}{3};2\right\}\)

\(A\cap B=\left\{2\right\}\) ; \(A\cup B=\left\{\dfrac{1}{3};\dfrac{1}{2};2\right\}\)

\(A\backslash B=\left\{\dfrac{1}{2}\right\}\) ; \(B\backslash A=\left\{\dfrac{1}{3}\right\}\)

7.

Các tập con:

\(\varnothing;\left\{a\right\};\left\{b\right\};\left\{c\right\};\left\{d\right\};\left\{a;b\right\};\left\{a;c\right\};\left\{a;d\right\};\left\{b;c\right\};\left\{b;d\right\};\left\{c;d\right\}\)

\(\left\{a;b;c\right\};\left\{a;b;d\right\};\left\{a;c;d\right\};\left\{b;c;d\right\};\left\{a;b;c;d\right\}\)

Giúp em báo cáo này đi ạ xoá hôg đc ạ 

Bài 5:

1: ΔABC vuông tại B

=>\(BA^2+BC^2=AC^2\)

=>\(AC^2=8^2+6^2=64+36=100=10^2\)

=>AC=10(cm)

Bài 1:

a: \(4x^2-4-\left(2x-1\right)\left(3x+4\right)=0\)

=>\(4x^2-4-\left(6x^2+8x-3x-4\right)=0\)

=>\(4x^2-4-\left(6x^2+5x-4\right)=0\)

=>\(4x^2-4-6x^2-5x+4=0\)

=>\(-2x^2-5x=0\)

=>-x(2x+5)=0

=>x(2x+5)=0

=>x∈{0;-5/2}

b: \(\frac{2x+1}{5}-\frac12=\frac{x-2}{6}-\frac{3-x}{4}\)

=>\(\frac{12\left(2x+1\right)}{60}-\frac{30}{60}=\frac{10\left(x-2\right)}{60}-\frac{15\left(3-x\right)}{60}\)

=>12(2x+1)-30=10(x-2)-15(3-x)

=>24x+12-30=10x-20-45+15x

=>24x-18=25x-65

=>24x-25x=-65+18

=>-x=-47

=>x=47

c: ĐKXĐ: x<>3; x<>-3

\(\frac{2}{x+3}=\frac{x-5}{x^2-9}-\frac{5}{3-x}\)

=>\(\frac{2}{x+3}-\frac{x-5}{\left(x-3\right)\left(x+3\right)}-\frac{5}{x-3}=0\)

=>\(\frac{2\left(x-3\right)-\left(x-5\right)-5\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=0\)

=>2(x-3)-(x-5)-5(x+3)=0

=>2x-6-x+5-5x-15=0

=>-4x-16=0

=>x=-4(nhận)

d: |x+2|=2x-3

=>\(\begin{cases}2x-3\ge0\\ \left(2x-3\right)^2=\left(x+2\right)^2\end{cases}\Rightarrow\begin{cases}x\ge\frac32\\ \left(2x-3-x-2\right)\left(2x-3+x+2\right)=0\end{cases}\)

=>\(\begin{cases}x\ge\frac32\\ \left(x-5\right)\left(3x-1\right)=0\end{cases}\)

=>x=5

PTHH : 2Al     +     6HCl  --> 2AlCl3   +    3H2 ↑   (1)

n_AlCl3 = \(\dfrac{m}{M}=\dfrac{13,35}{27+35,5.3}=0.1\left(mol\right)\) 

Từ (1) => n_HCl   =   2n_H2  = 0.2 (mol)

=> m_HCl = n.M  =  0.2 x  36.5 = 7.3 (g)

\(PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\\ n_{AlCl_3}=\dfrac{m}{M}=\dfrac{13,35}{133,5}=0,1\left(mol\right)\\ Theo.PTHH:n_{HCl}=3.n_{AlCl_3}=3.0,1=0,3\left(mol\right)\\ m_{HCl}=n.M=0,3.36,5=10,95\left(g\right)\)

Bài 3 :

a) n_KMnO4 = 31,6/158 = 0,2 (mol)

PTHH : \(2KMnO_4-t^o->K_2MnO_4+MnO_2+O_2\)

Theo pthh : nO2 = 1/2 nKMnO4 = 0,1 (mol)

=> V_O2 = 0,1.22,4 = 2,24 (l)

b) PTHH : \(2Mg+O_2-t^o->2MgO\)

Theo pthh : nMgO = 2nO2 = 0,2 (mol)

=> m_MgO = 0,2.40 = 8 (g)

Bài 5 :

a) nCu = 0,2 (mol)

nO2 = 0,2 (mol)

PTHH : \(2Cu+O_2-t^o->2CuO\)

Nhận thấy : \(\dfrac{nCu}{2}< nO_2\left(\dfrac{0,2}{2}< 0,2\right)\) => Spu,Cu hết, O2 dư

Theo pthh : nCuO = nCu = 0,2 (mol)

=> mCuO = 0,2.80 = 16 (g)

b) PTHH : \(CuO+H_2-t^o->Cu+H_2O\)

Theo pthh : nH2 = nCuO = 0,2 (mol)

=> V_H2 = 0,2.22,4 = 4,48 (l)

Part 1:

1. with

2. does

3. after

4. if

5. where

6. I won't

Part 2:

1B 2F 3A 4E 5C 6D

Part 3:

1. invited

2. will play

3. have just won

4. be cleaned

5. to take

6. beautifully

7. impression

Part 4:

1. read -> reading

2. disappointing -> disappointed

3. environment -> environmental

4. who -> that

Part 5:

1. the bad weather

2. are made to study

3. were good at learning

4. going to the English

hic cíu mng oi

a: ĐKXĐ: \(x\notin\left\{10;-10;\sqrt{10};-\sqrt{10}\right\}\)

b: \(A=\dfrac{5x^3+50x+2x^2+20+5x^3-50x-2x^2+20}{\left(x^2-10\right)\left(x^2+10\right)}\cdot\dfrac{x^2-100}{x^2+4}\)

\(=\dfrac{10x^3+40}{\left(x^2-10\right)\left(x^2+10\right)}\cdot\dfrac{x^2-100}{x^2+4}\)