bài nào zậy bạn

Câu 3 và caau4 bài giải phương trình nhé

\(A=\dfrac{1}{5}+\dfrac{1}{10}+\dfrac{1}{20}+\dfrac{1}{40}+\dfrac{1}{80}+\dfrac{1}{160}\\ A=\dfrac{35}{160}+\dfrac{16}{160}+\dfrac{8}{160}+\dfrac{4}{160}+\dfrac{2}{160}+\dfrac{1}{160}\\ A=\dfrac{66}{160}\\ A=\dfrac{33}{80}.\)

e cam on ah !

b: \(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)

\(=\left(x^2+x\right)^2+3\left(x^2+x\right)+2-12\)

\(=\left(x^2+x\right)^2+3\left(x^2+x\right)-10\)

\(=\left(x^2+x+5\right)\left(x^2+x-2\right)\)

\(=\left(x^2+x+5\right)\left(x+2\right)\left(x-1\right)\)

c: \(\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)

\(=\left(x^2+8x\right)^2+22\left(x^2+8x\right)+105+15\)

\(=\left(x^2+8x\right)^2+22\left(x^2+8x\right)+120\)

\(=\left(x^2+8x+12\right)\left(x^2+8x+10\right)\)

\(=\left(x+2\right)\left(x+6\right)\left(x^2+8x+10\right)\)

Câu 8  :B

m NaCl = 200.5% = 10(gam)

m H2O = m dd - m NaCl = 200 - 10 = 190(gam) 

Câu 8:

mNaCl=5%.200=10(g)

=>mH2O=mddNaCl - mNaCl= 200-10=190(g)

=> Điều chế: Hòa tan 10 gam NaCl vào 190 gam H2O

=> CHỌN B

1 D

2 C

3 B

4 A

5 B

6 C

1 D

2 C

3 B

4 A

5 B

6 C

\(3,8276< \overline{3,8ab5}< 3,836\)

=>\(276< \overline{ab5}< 360\)

=>\(\left(a,b\right)\in\left\{\left(2;8\right);\left(2;9\right);\left(3;0\right);\left(3;1\right);\left(3;2\right);\left(3;3\right);\left(3;4\right);\left(3;5\right)\right\}\)