Câu 5: 

\(\Leftrightarrow-x^2+7x-9+2x-9=0\)

\(\Leftrightarrow x^2-9x+18=0\)

=>x=3

=>Chọn A

không hiểu chỗ nào?

?

Bài 5:

a: 2x-(3-5x)=4(x+3)

=>2x-3+5x=4x+12

=>7x-3=4x+12

=>3x=15

=>x=5

b: =>5/3x-2/3+x=1+5/2-3/2x

=>25/6x=25/6

=>x=1

c: 3x-2=2x-3

=>3x-2x=-3+2

=>x=-1

d: =>2u+27=4u+27

=>u=0

e: =>5-x+6=12-8x

=>-x+11=12-8x

=>7x=1

=>x=1/7

f: =>-90+12x=-45+6x

=>12x-90=6x-45

=>6x-45=0

=>x=9/2

PTHH: \(Zn+S\underrightarrow{t^o}ZnS\)

Từ đề bài, ta có: \(\left\{{}\begin{matrix}n_S=n_{ZnS}=0,15\left(mol\right)\\n_{Zn\left(dư\right)}=0,05\left(mol\right)\end{matrix}\right.\)

c) PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)

                \(ZnS+2HCl\rightarrow ZnCl_2+H_2S\uparrow\)

Theo PTHH: \(\Sigma n_{HCl}=2n_{ZnS}+2n_{Zn\left(dư\right)}=0,4\left(mol\right)\)

\(\Rightarrow m_{ddHCl}=\dfrac{0,4\cdot36,5}{20\%}=73\left(g\right)\)

nhìn hơi rối

bn tách ra đc k ặc

a,1/5+4/11+4/5+7/11

=(1/5+4/5)+(4/11+7/11)

=1+1

=2

Chọn B

1367.54+1367.45+1367

=1367.(54+45+1)
=1367.100

=136700

a) 1/5 + 4/11 + 4/5 + 7/11 = (1/5 + 4/5) + (4/11 + 7/11) = 1 + 1 = 2

=> B

b) 1367 x 54 + 1367 x 45 + 1367 = 1367 x (54 + 45 + 1) = 1367 x 100 = 136700

ĐKXĐ: \(\left\{{}\begin{matrix}2x+5>=0\\4-2x>=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x>=-5\\2x< =4\end{matrix}\right.\Leftrightarrow-\dfrac{5}{2}< =x< =2\)

\(x^2+\sqrt{2x+5}+\sqrt{4-2x}=4x-1\)

=>\(x^2-4+\sqrt{2x+5}-3+\sqrt{4-2x}=4x-1-7\)

=>\(\left(x-2\right)\left(x+2\right)+\dfrac{2x+5-9}{\sqrt{2x+5}+3}+\sqrt{4-2x}=4x-8\)

=>\(\left(x-2\right)\left[\left(x+2\right)+\dfrac{2}{\sqrt{2x+5}+3}-4\right]+\sqrt{4-2x}=0\)

=>\(-\left(2-x\right)\left[\left(x-2\right)+\dfrac{2}{\sqrt{2x+5}+3}\right]+\sqrt{2\left(2-x\right)}=0\)

=>\(\sqrt{2-x}\left[-\sqrt{2-x}\left(x-2+\dfrac{2}{\sqrt{2x+5}+3}\right)+\sqrt{2}\right]=0\)

=>\(\sqrt{2-x}=0\)

=>x=2(nhận)

20 < 5x < 45

=> 5 . 4 < 5x < 5 . 9

=> 4 < x < 9

=> \(x\in\left\{5;6;7;8\right\}\)

\(20< 5x< 40\)

\(20=5.4;40=5.8\)

\(\Rightarrow\text{A=\text{{}5;6;7}}\)

Lỗi tí thông cảm

A={ 5 ; 6 ; 7 ; 8 }