a)(5³)³.(5²)⁴=5⁹.5⁸=5¹⁷

b)(4²)².(4³)⁵=4⁴.4¹⁵=4¹⁹

^{kick mik nha}

\(125^3.25^4=25^9.25^4=25^{13}\)

\(16^2.64^5=4^4.4^{15}=4^{19}\)

d) Gọi x,y lần lượt là số mol Al, Fe

\(\left\{{}\begin{matrix}27x+56y=8,3\\1,5x+y=0,25\end{matrix}\right.\)

=> x=0,1 ; y=0,1

Kết tủa : Al(OH)3, Fe(OH)2 

Bảo toàn nguyên tố Al: \(n_{Al\left(OH\right)_3}=n_{Al}=0,1\left(mol\right)\)

Bảo toàn nguyên tố Fe: \(n_{Fe\left(OH\right)_2}=n_{Fe}=0,1\left(mol\right)\)

=> \(m=0,1.78+0,1.90=16,8\left(g\right)\)

Nung kết tủa thu được chất rắn : Al2O3 và FeO

Bảo toàn nguyên tố Al: \(n_{Al_2O_3}.2=n_{Al}\Rightarrow n_{Al_2O_3}=0,05\left(mol\right)\)

Bảo toàn nguyên tố Fe: \(n_{FeO}=n_{Fe}=0,1\left(mol\right)\)

=> \(a=0,05.102+0,1.72=12,3\left(g\right)\)

1: A=-1/2*xy^3*4x^2y^2=-2x^3y^5

Bậc là 8

Phần biến là x^3;y^5

Hệ số là -2

2:

a: P(x)=3x+4x^4-2x^3+4x^2-x^4-6

=3x^4-2x^3+4x^2+3x-6

Q(x)=2x^4+4x^2-2x^3+x^4+3

=3x^4-2x^3+4x^2+3

b: A(x)=P(x)-Q(x)

=3x^4-2x^3+4x^2+3x-6-3x^4+2x^3-4x^2-3

=3x-9

A(x)=0

=>3x-9=0

=>x=3

Câu 4: ĐKXĐ: x>=1/2

Ta có: \(2\left(x-\sqrt{2x^2+5x-3}\right)=1+x\left(\sqrt{2x-1}-2\sqrt{x+3}\right)\)

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

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

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

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

=>\(\left(\sqrt{2x-1}-\sqrt{4x+12}\right)\left(\sqrt{2x-1}-x\right)=0\)

TH1: \(\sqrt{2x-1}-\sqrt{4x+12}=0\)

=>\(\sqrt{2x-1}=\sqrt{4x+12}\)

=>4x+12=2x-1

=>2x=-13

=>\(x=-\frac{13}{2}\) (loại)

TH2: \(\sqrt{2x-1}-x=0\)

=>\(\sqrt{2x-1}=x\)

=>\(\begin{cases}2x-1=x^2\\ x\ge0\end{cases}\Rightarrow\begin{cases}x^2-2x+1=0\\ x\ge0\end{cases}\Rightarrow\begin{cases}\left(x-1\right)^2=0\\ x\ge0\end{cases}\)

=>x-1=0

=>x=1(nhận)

Bài 3: 

a: Gọi OK là khoảng cách từ O đến AB

Suy ra: K là trung điểm của AB

hay \(AK=BK=\dfrac{AB}{2}=\dfrac{8}{2}=4\left(cm\right)\)

Áp dụng định lí Pytago vào ΔOKA vuông tại K, ta được:

\(OA^2=OK^2+KA^2\)

hay OK=3(cm)

Câu 5:

1: cos3x-sin 3x=-1

=>\(\sin3x-cos3x=1\)

=>\(\sqrt2\cdot\sin\left(3x-\frac{\pi}{4}\right)=1\)

=>\(\sin\left(3x-\frac{\pi}{4}\right)=\frac{1}{\sqrt2}\)

=>\(\left[\begin{array}{l}3x-\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\ 3x-\frac{\pi}{4}=\pi-\frac{\pi}{4}+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}3x=\frac{\pi}{2}+k2\pi\\ 3x=\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\pi}{6}+\frac{k2\pi}{3}\\ x=\frac{\pi}{3}+\frac{k2\pi}{3}\end{array}\right.\)

2: \(\sqrt3\cdot\sin\left(\frac{x}{2}\right)-cos\left(\frac{x}{2}\right)-\sqrt2=0\)

=>\(\sqrt3\cdot\sin\left(\frac{x}{2}\right)-cos\left(\frac{x}{2}\right)=\sqrt2\)

=>\(\frac{\sqrt3}{2}\cdot\sin\left(\frac{x}{2}\right)-\frac12\cdot cos\left(\frac{x}{2}\right)=\frac{\sqrt2}{2}\)

=>\(\sin\left(\frac{x}{2}-\frac{\pi}{6}\right)=\sin\left(\frac{\pi}{4}\right)\)

=>\(\left[\begin{array}{l}\frac{x}{2}-\frac{\pi}{6}=\frac{\pi}{4}+k2\pi\\ \frac{x}{2}-\frac{\pi}{6}=\pi-\frac{\pi}{4}+k2\pi=\frac34\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}\frac{x}{2}=\frac{\pi}{6}+\frac{\pi}{4}+k2\pi=\frac{5}{12}\pi+k2\pi\\ \frac{x}{2}=\frac34\pi+\frac{\pi}{6}+k2\pi=\frac{11}{12}\pi+k2\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}x=\frac56\pi+k4\pi\\ x=\frac{11}{6}\pi+k4\pi\end{array}\right.\)

3: 3*sin 4x+4* cos4x=5

=>\(\frac35\cdot\sin4x+\frac45\cdot cos4x=1\)

=>\(\sin\left(4x+\alpha\right)=1\)

=>\(4x+\alpha=\frac{\pi}{2}+k2\pi\)

=>\(4x=\frac{\pi}{2}-\alpha+k2\pi\)

=>\(x=\frac{\pi}{8}-\frac{\alpha}{4}+\frac{k\pi}{2}\)

Bài 4:

1: \(3\cdot\sin^23x-4\cdot\sin3x+1=0\)

=>\(3\cdot\sin^23x-3\cdot\sin3x-\sin3x+1=0\)

=>(sin 3x-1)(3sin 3x-1)=0

TH1: sin 3x-1=0

=>sin 3x=1

=>\(3x=\frac{\pi}{2}+k2\pi\)

=>\(x=\frac{\pi}{6}+\frac{k2\pi}{3}\)

TH2: 3 sin 3x-1=0

=>3sin 3x=1

=>sin 3x=1/3

=>\(\left[\begin{array}{l}3x=\arcsin\left(\frac13\right)+k2\pi\\ 3x=\pi-\arcsin\left(\frac13\right)+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac13\cdot\arcsin\left(\frac13\right)+\frac{k2\pi}{3}\\ x=\frac{\pi}{3}-\frac13\cdot\arcsin\left(\frac13\right)+\frac{k2\pi}{3}\end{array}\right.\)

2: \(4\cdot cos^2\left(\frac{x}{2}\right)-1=0\)

=>\(4\cdot cos^2\left(\frac{x}{2}\right)=1\)

=>\(cos^2\left(\frac{x}{2}\right)=\frac14\)

=>\(\left[\begin{array}{l}cos\left(\frac{x}{2}\right)=\frac12\\ cos\left(\frac{x}{2}\right)=-\frac12\end{array}\right.\Rightarrow\left[\begin{array}{l}\frac{x}{2}=\frac{\pi}{3}+k2\pi\\ \frac{x}{2}=-\frac{\pi}{3}+k2\pi\\ \frac{x}{2}=\frac23\pi+k2\pi\\ \frac{x}{2}=-\frac23\pi+k2\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}x=\frac{2\pi}{3}+k4\pi\\ x=-\frac23\pi+k4\pi\\ x=\frac43\pi+k4\pi\\ x=-\frac43\pi+k4\pi\end{array}\right.\)

3: \(3\cdot\tan^24x-\sqrt3\cdot\tan4x=0\)

=>\(\sqrt3\cdot\tan4x\left(\sqrt3\cdot\tan4x-1\right)=0\)

TH1: tan 4x=0

=>\(4x=k\pi\)

=>\(x=\frac{k\pi}{4}\)

TH2: \(\sqrt3\cdot\tan4x-1=0\)

=>\(\tan4x=\frac{1}{\sqrt3}\)

=>\(4x=\frac{\pi}{6}+k\pi\)

=>\(x=\frac{\pi}{24}+\frac{k\pi}{4}\)

5: \(\sin^2x+cosx-1=0\)

=>\(1-cos^2x+cosx-1=0\)

=>\(-cos^2x+cosx=0\)

=>cosx(cosx-1)=0

TH1: cosx=0

=>\(x=\frac{\pi}{2}+k\pi\)

TH2: cos x-1=0

=>cosx =1

=>\(x=k2\pi\)

6: \(\cot^22x-2\cdot\cot2x-3=0\)

=>(cot 2x-3)(cot 2x+1)=0

TH1: cot 2x-3=0

=>cot 2x=3

=>\(2x=arc\cot\left(3\right)+k\pi\)

=>\(x=\frac12\cdot arc\cot\left(3\right)+\frac{k\pi}{2}\)

TH2: cot 2x+1=0

=>cot 2x=-1

=>\(2x=-\frac{\pi}{4}+k\pi\)

=>\(x=-\frac{\pi}{8}+\frac{k\pi}{2}\)