Ta có: 

⇔x=177

a)36x²-y²=(6x-y)(6x+y)

b)25x²+10xy+y²=(5x+y)²

c)9x²+24xy+16y²=(3x+4y)²

d)4x²-9y⁴=(2x-3y²)(2x+3y²)

e)x³-8=(x-8)(x²+2x+4)

f)\(x^3-\dfrac{y^3}{27}=\left(x-\dfrac{y}{3}\right)\left(x^2+x\dfrac{y}{3}+\dfrac{y^2}{9}\right)\)

comfortably

happily

428=2².107

422=2.211

115=5.23

180=2².3².5

160=2⁵.5

190=2.5.9

250=2.5³

350=2.5².7

324=2².3⁴

364=2².7.13

270=2.3³.5

290=2.5.29

120=2³.3.5

150=2.3.5²

160=2⁵.5

ĐK:\(x\ge2\)

ĐK:\(x\ge2\)

Ta có:\(\sqrt{x^2-4x+4}=-x+4\)

     \(\Leftrightarrow\sqrt{\left(x-2\right)^2}=-x+4\)

     \(\Leftrightarrow\left|x-2\right|=-x+4\)

     \(\Leftrightarrow\left[{}\begin{matrix}x-2=-x+4\\x-2=x-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\0=-2\left(loại\right)\end{matrix}\right.\Leftrightarrow x=3\)

\(m_{HCl}=\dfrac{14,6.80}{100}=11,68\left(g\right)\Rightarrow n_{HCl}=\dfrac{11,68}{36,5}=0,32\left(mol\right)\);

\(n_{FeO}=\dfrac{7,2}{72}=0,1\left(mol\right)\)

\(PTHH:FeO+2HCl\rightarrow FeCl_2+H_2O\)

Mol:          0,1        0,2         0,1

Ta có tỉ lệ:\(\dfrac{0,32}{2}>\dfrac{0,1}{1}\) =>HCl dư,FeO phản ứng hết

a)m_{HCl dư}=(0,32-0,2).36,5=4,38 (g)

b)\(m_{FeCl_2}=0,1.127=12,7\left(g\right)\)

mik viết nhầm,là has given

1.A

2.A

3.D

4.C

5.ko có đáp án đúng

6.B

7.D

8.A

9.C

10.D