Sửa đề là : 4.6 (g) 

\(n_{H_2}=\dfrac{2.24}{22.4}=0.1\left(mol\right)\)

\(A+H_2O\rightarrow AOH+\dfrac{1}{2}H_2\)

\(0.2...............................0.1\)

\(M_A=\dfrac{4.6}{0.2}=23\left(\dfrac{g}{mol}\right)\)

\(A:Na\)

Đề này C1 em sửa thành 4,6 gam kim loại như bạn dưới, C2 em sửa thành 22,4 lít H₂

a: \(\left|\overrightarrow{AB}-\overrightarrow{AC}\right|=\left|\overrightarrow{CB}\right|=10a\)

b: \(\left|\overrightarrow{AB}+\overrightarrow{AC}\right|=\dfrac{BC}{2}=5a\)

1) Vì x=25 thỏa mãn ĐKXĐ nên Thay x=25 vào biểu thức \(A=\dfrac{\sqrt{x}-2}{x+1}\), ta được:

\(A=\dfrac{\sqrt{25}-2}{25+1}=\dfrac{5-2}{25+1}=\dfrac{3}{26}\)

Vậy: Khi x=25 thì \(A=\dfrac{3}{26}\)

2) Ta có: \(B=\dfrac{\sqrt{x}-3}{\sqrt{x}+1}+\dfrac{2x+8\sqrt{x}-6}{x-\sqrt{x}-2}\)

\(=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}+\dfrac{2x+8\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x-5\sqrt{x}+6+2x+8\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)

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

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

\(=\dfrac{3\sqrt{x}}{\sqrt{x}-2}\)

câu 3 chứ

Câu 1: Vì (d') vuông góc với (d) nên \(a\cdot\dfrac{-1}{3}=-1\)

hay a=3

Vậy: (d'): y=3x+b

Thay x=4 và y=-5 vào (d'), ta được:

b+12=-5

hay b=-17

Bài 6:

a) \(x^2-2x+4=\left(x^2-2x+1\right)+3=\left(x-1\right)^2+3>0\forall x\)

b) \(-x^2+4x-5=-\left(x^2-4x+4\right)-1=-\left(x-2\right)^2-1< 0\forall x\)

c) \(\left(x-2\right)\left(x-4\right)+3=x^2-6x+11=\left(x^2-6x+9\right)+2=\left(x-3\right)^2+2>0\forall x\)

d) \(-2x^2+5x-19=\dfrac{-4x^2+10x-38}{2}=\dfrac{-\left(4x^2-10x+6,25\right)-31,75}{2}=\dfrac{-\left(2x-2,5\right)^2-31,75}{2}< 0\forall x\)

Câu 4:

a) \(x^5-x^3-x^2+1=\left(x^5-x^3\right)-\left(x^2-1\right)=x^3\left(x^2-1\right)-\left(x-1\right)\left(x+1\right)=x^3\left(x-1\right)\left(x+1\right)-\left(x-1\right)\left(x+1\right)=\left(x-1\right)\left(x+1\right)\left(x^3-1\right)=\left(x-1\right)^2\left(x+1\right)\left(x^2+x+1\right)\)

Câu 5:

\(a^3+b^3=3ab-1\\ \Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)-3ab+1=0\\ \Leftrightarrow\left(a+b+1\right)\left(a^2+2ab+b^2-a-b+1\right)-3ab\left(a+b+1\right)=0\\ \Leftrightarrow\left(a+b+1\right)\left(a^2+b^2+1-ab-a-b\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a+b+1=0\left(vô.lí.do.a,b>0\right)\\a^2+b^2+1-ab-a-b=0\left(1\right)\end{matrix}\right.\\ \left(1\right)\Leftrightarrow2a^2+2b^2+2-2ab-2a-2b=0\\ \Leftrightarrow\left(a-b\right)^2+\left(a-1\right)^2+\left(b-1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}a-b=0\\a-1=0\\b-1=0\end{matrix}\right.\Leftrightarrow a=b=1\)

Vậy \(T=\left(1-2\right)^{2020}+\left(1-1\right)^{2021}=\left(-1\right)^{2020}+0=1\)

Câu 1: 

TXĐ: D=R

\(f\left(-x\right)=2\cdot\left(-x\right)^4-3\cdot\left(-x\right)^2+1=2x^4-3x^2+1=f\left(x\right)\)

Vậy: f(x) là hàm số chẵn

1.

\(\Leftrightarrow\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)=0\)

\(\Leftrightarrow sin\left(x-\dfrac{\pi}{4}\right)=0\)

\(\Leftrightarrow x-\dfrac{\pi}{4}=k\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{4}+k\pi\)

2.

\(\Leftrightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=1\)

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{4}=\dfrac{\pi}{4}+k2\pi\\x+\dfrac{\pi}{4}=\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)

3.

\(\Leftrightarrow\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x=\dfrac{5}{8}\)

\(\Leftrightarrow1-\dfrac{1}{2}sin^22x=\dfrac{5}{8}\)

\(\Leftrightarrow1-\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{2}cos4x\right)=\dfrac{5}{8}\)

\(\Leftrightarrow\dfrac{3}{4}+\dfrac{1}{4}cos4x=\dfrac{5}{8}\)

\(\Leftrightarrow cos4x=-\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}4x=\dfrac{2\pi}{3}+k2\pi\\4x=-\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k\pi}{2}\\x=-\dfrac{\pi}{6}+\dfrac{k\pi}{2}\end{matrix}\right.\)

4.

\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=\dfrac{1}{4}\)

\(\Leftrightarrow1-\dfrac{3}{4}sin^22x=\dfrac{1}{4}\)

\(\Leftrightarrow1-\dfrac{3}{4}\left(\dfrac{1}{2}-\dfrac{1}{2}cos4x\right)=\dfrac{1}{4}\)

\(\Leftrightarrow cos4x=-1\)

\(\Leftrightarrow4x=\pi+k2\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)

a: Tỉ số là 3/2

b: Tỉ số phần trăm là;

40/(30+40+20+20+5)=34,78%

Bài 1.2

1: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)

2) Ta có: \(A=\dfrac{2\sqrt{x}}{\sqrt{x}+3}-\dfrac{\sqrt{x}+1}{3-\sqrt{x}}-\dfrac{3-11\sqrt{x}}{x-9}\)

\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{3-11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{2x-6\sqrt{x}+x+4\sqrt{x}+3-3+11\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{3x+9\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}-3}\)