1.

\(cos^2x-\sqrt{3}sin2x=1+sin^2x\)

\(\Leftrightarrow cos2x-\sqrt{3}sin2x=1\)

\(\Leftrightarrow\dfrac{1}{2}cos2x-\dfrac{\sqrt{3}}{2}sin2x=\dfrac{1}{2}\)

\(\Leftrightarrow cos\left(2x+\dfrac{\pi}{3}\right)=\dfrac{1}{2}\)

\(\Leftrightarrow2x+\dfrac{\pi}{3}=\pm\dfrac{\pi}{3}+k2\pi\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=-\dfrac{\pi}{3}+k\pi\end{matrix}\right.\)

2.

\(10cos^2x-5sinx.cosx+3sin^2x=4\)

\(\Leftrightarrow20cos^2x-10sinx.cosx+6sin^2x=8\)

\(\Leftrightarrow20cos^2x-10-10sinx.cosx+6sin^2x-3=-5\)

\(\Leftrightarrow7cos2x-5sin2x=-5\)

\(\Leftrightarrow\sqrt{74}\left(\dfrac{7}{\sqrt{74}}cos2x-\dfrac{5}{\sqrt{74}}sin2x\right)=-5\)

\(\Leftrightarrow cos\left(2x+arccos\dfrac{7}{\sqrt{74}}\right)=-\dfrac{5}{\sqrt{74}}\)

\(\Leftrightarrow2x+arccos\dfrac{7}{\sqrt{74}}=\pm arccos\dfrac{5}{\sqrt{74}}+k2\pi\)

\(\Leftrightarrow x=-\dfrac{1}{2}arccos\dfrac{7}{\sqrt{74}}\pm\dfrac{1}{2}arccos\dfrac{5}{\sqrt{74}}+k\pi\)

\(1,=3ab\left(1-2a+b\right)\\ 2,=\left(x-y\right)\left(x+y\right)-7\left(x+y\right)=\left(x+y\right)\left(x-y-7\right)\\ 3,=\left(a-5\right)\left(5a-2\right)\\ 4,=5x\left(x-3\right)-\left(x-3\right)\left(x+3\right)=\left(x-3\right)\left(4x-3\right)\\ 5,=9a^2-\left(b-2\right)^2=\left(3a-b+2\right)\left(3a+b-2\right)\\ 6,=2x^2-4x+3x-6=\left(x-2\right)\left(2x+3\right)\\ 7,=3x^2\left(2x-5\right)\\ 8,=\left(3x-5\right)\left(3x+5\right)\\ 9,=4x^2\left(x-y\right)-x\left(x-y\right)=x\left(4x-1\right)\left(x-y\right)\)

Câu 58: B

Câu 59: C

2) nH2=0,1(mol)

a) PTHH: Fe +2  HCl -> FeCl2 + H2

0,1______0,2______0,1____0,1(mol)

nFe=nH2=0,1(mol)

=>mFe=nFe.M(Fe)=0,1.56=5,6(g)

=> mFeO=mX-mFe= 9,2-5,6=3,6(g)

=> nFeO=mFeO/M(FeO)=3,6/72=0,05(mol)

PTHH: FeO +2 HCl -> FeCl2 + H2

0,05_________0,1___0,05__0,05(mol)

b) Sao lại mỗi oxit a, có một oxit thôi mà :(  Chắc % KL mỗi chất.

%mFeO=(mFeO/mhh).100%=(3,6/9,2).100=39,13%

=>%mFe=100%-%mFeO=100%-39,13%=60,87%

c) nHCl(tổng)= 2.nFe +2.nFeO=2.0,1+2.0,05=0,3(mol)

=>mHCl=nHCl.M(HCl)=0,3.36,5=10,95(g)

=>mddHCl=(mHCl.100%/C%ddHCl=(10,95.100)/7,3=150(g)

d) - Dung dich thu được chứa FeCl2.

mFeCl2=nFeCl2(tổng) . M(FeCl2)= (0,1+0,05).127=19,05(g)

mddFeCl2=mddHCl+mhh-mH2=150+9,2-0,1.2=159(g)

=> C%ddFeCl2=(mFeCl2/mddFeCl2).100%=(19,05/159).100=11,981%

BPT đã cho vô nghiệm khi và chỉ khi BPT \(f\left(x\right)\le0\) nghiệm đúng với mọi x

TH1: \(\left\{{}\begin{matrix}2m^2+m-6=0\\2m-3=0\end{matrix}\right.\) \(\Rightarrow m=\dfrac{3}{2}\)

TH2: \(\left\{{}\begin{matrix}2m^2+m-6< 0\\\Delta=\left(2m-3\right)^2+4\left(2m^2+m-6\right)\le0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2m^2+m-6< 0\\12m^2-8m-15\le0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-3< m< \dfrac{3}{2}\\-\dfrac{5}{6}\le m\le\dfrac{3}{2}\end{matrix}\right.\) \(\Rightarrow-\dfrac{5}{6}\le m< \dfrac{3}{2}\)

Kết hợp 2 trường hợp ta được \(-\dfrac{5}{6}\le m\le\dfrac{3}{2}\)

\(\Leftrightarrow\left(x+3\right)\sqrt{2x^2+1}-\left(x+3\right)=x^2\)

=>\(\left(x+3\right)\cdot\left(\sqrt{2x^2+1}-1\right)=x^2\)

=>\(\left(x+3\right)\cdot\dfrac{2x^2+1-1}{\sqrt{2x^2+1}+1}-x^2=0\)

=>\(x^2\left(\dfrac{2\left(x+3\right)}{\sqrt{2x^2+1}+1}-1\right)=0\)

=>x^2=0 hoặc \(\dfrac{2\left(x+3\right)}{\sqrt{2x^2+1}+1}=1\)

=>\(\left[{}\begin{matrix}x=0\\\sqrt{2x^2+1}+1=2x+6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\2x^2+1=\left(2x+5\right)^2;x>=-\dfrac{5}{2}\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=0\\4x^2+20x+25-2x^2-1=0;x>=-\dfrac{5}{2}\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=0\\\left\{{}\begin{matrix}2x^2+20x+24=0\\x>=-\dfrac{5}{2}\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5+\sqrt{13}\end{matrix}\right.\)

=>Phương trình này có 2 nghiệm

Ta có:

     (2 - 3x)(x + 8) = (3x - 2)(3 - 5x)

⇔ (2 - 3x)(x + 8) - (3x - 2)(3 - 5x) = 0

⇔ (2 - 3x)(x + 8) + (2 - 3x)(3 - 5x) = 0

⇔ (2 - 3x)(x + 8 + 3 - 5x) = 0

⇔ (2 - 3x)(11 - 4x) = 0

⇔ 2 - 3x = 0 hay 11 - 4x = 0

⇔ 2 = 3x hay 11 = 4x

⇔ x = \(\dfrac{2}{3}\) hay x = \(\dfrac{11}{4}\)

Vậy tập nghiệm của pt S = \(\left\{\dfrac{2}{3};\dfrac{11}{4}\right\}\)

khum bt mới lớp 5 hoi à(k10)

<=> (2-3x ) (x+8) + (2-3x ) (3-5x)=0
<=> (2-3x ) ( x+8 +  3-5x ) =0 
<=> (2-3x ) ( 11 - 4x ) = 0
 => 2-3x  =0 hoặc 11-4x =0  
       3x = 2            4x =11
         x = 2/3         x    = 11/4

Mình làm tròn nha!

0,(9)\(\approx\)1 =\(\frac{1}{1}\)

1=1/1