Bài 1:

Thời gian ô tô đi hết quãng đường:

\(t=\dfrac{S}{v}=\dfrac{60}{40}=\dfrac{3}{2}\left(h\right)\)

Vận tốc lúc về là:

\(40+10=50\left(\dfrac{km}{h}\right)\)

Thời gian ô tô đi về:

\(t_2=\dfrac{S_2}{v_2}=\dfrac{60}{50}=\dfrac{6}{5}\left(h\right)\)

a, \(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

PTHH: Fe + H₂SO₄ → FeSO₄ + H₂

Mol:     0,25    0,25          0,25     0,25

\(m_{Fe}=0,25.56=14\left(g\right)\)

\(m_{H_2SO_4}=0,25.98=24,5\left(g\right)\)

b, \(m_{FeSO_4}=0,25.162=40,5\left(g\right)\)

khó thật , đọc hỉu j đâu 

Bn ơi!Nếu bài này có hình vẽ thì bn đăng luôn hình nha!

\(10,\\ a^2+b^2+c^2+d^2+e^2\ge a\left(b+c+d+e\right)\\ \Leftrightarrow4a^2+4b^2+4c^2+4d^2+4e^2\ge4ab+4ac+4ad+4ae\\ \Leftrightarrow\left(a^2-4ab+4b^2\right)+\left(a^2-4ac+4c^2\right)+\left(a^2-4ad+4d^2\right)+\left(a^2-4ae+4e^2\right)\ge0\\ \Leftrightarrow\left(a-2b\right)^2+\left(a-2c\right)^2+\left(a-2d\right)^2+\left(a-2e\right)^2\ge0\left(luôn.đúng\right)\)

Dấu \("="\Leftrightarrow\dfrac{a}{2}=b=c=d=e\)

\(4,\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac-\dfrac{1}{4}\left(2a^2+2b^2+2c^2-2ab-2ac-2bc\right)\ge3ab+3bc+3ca\\ \Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac-\dfrac{1}{2}a^2-\dfrac{1}{2}b^2-\dfrac{1}{2}c^2-ab-bc-ac\ge0\\ \Leftrightarrow\dfrac{1}{2}a^2+\dfrac{1}{2}b^2+\dfrac{1}{2}c^2+ab+ac+bc\ge0\\ \Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac\ge0\\ \Leftrightarrow\left(a+b+c\right)^2\ge0\left(luôn.đúng\right)\)

Dấu \("="\Leftrightarrow a+b+c=0\)

\(1,a^2+b^2+c^2\ge ab+bc+ca\\ \Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc\ge0\\ \Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ac+a^2\right)\ge0\\ \Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\left(luôn.đúng\right)\)

Dấu \("="\Leftrightarrow a=b=c\)

\(2,a^2+b^2+c^2+3\ge2\left(a+b+c\right)\\ \Leftrightarrow\left(a^2-2a+1\right)+\left(b^2-2b+1\right)+\left(c^2-2c+1\right)\ge0\\ \Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2\ge0\left(luôn.đúng\right)\)

Dấu \("="\Leftrightarrow a=b=c=1\)

\(3,\Leftrightarrow\dfrac{a^2+b^2+c^2}{3}\ge\dfrac{\left(a+b+c\right)^2}{9}\\ \Leftrightarrow9\left(a^2+b^2+c^2\right)\ge3\left(a+b+c\right)^2\\ \Leftrightarrow3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\\ \Leftrightarrow3a^2+3b^2+3c^2-a^2-b^2-c^2-2ab-2ac-2bc\ge0\\ \Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc\ge0\left(luôn.đúng.do.câu.1\right)\)

Dấu \("="\Leftrightarrow a=b=c\)

\(4,\Leftrightarrow a^2b^2+b^2c^2+a^2c^2+2a^2bc+2ab^2c+2abc^2\ge3a^2bc+3ab^2c+3abc^2\\ \Leftrightarrow a^2b^2+b^2c^2+a^2c^2-a^2bc-ab^2c-abc^2\ge0\\ \Leftrightarrow2a^2b^2+2b^2c^2+2a^2c^2-2a^2bc-2ab^2c-2abc^2\ge0\\ \Leftrightarrow\left(a^2b^2-2a^2bc+a^2c^2\right)+\left(a^2b^2-2ab^2c+b^2c^2\right)+\left(b^2c^2-2abc^2+a^2c^2\right)\ge0\\ \Leftrightarrow\left(ab-ac\right)^2+\left(ab-bc\right)^2+\left(bc-ac\right)^2\ge0\left(luôn.đúng\right)\)

Dấu \("="\Leftrightarrow ab=bc=ac\)

a: |x|=5,6

=>\(\left[{}\begin{matrix}x=5,6\\x=-5,6\end{matrix}\right.\)

c: \(\left|x\right|=3\dfrac{1}{5}\)

=>\(\left|x\right|=3,2\)

=>\(\left[{}\begin{matrix}x=3,2\\x=-3,2\end{matrix}\right.\)

d: |x|=-2,1

mà -2,1<0

nên \(x\in\varnothing\)

d: |x-3,5|=5

=>\(\left[{}\begin{matrix}x-3,5=5\\x-3,5=-5\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=8,5\\x=-1,5\end{matrix}\right.\)

e: \(\left|x+\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)

=>\(\left|x+\dfrac{3}{4}\right|=\dfrac{1}{2}\)

=>\(\left[{}\begin{matrix}x+\dfrac{3}{4}=\dfrac{1}{2}\\x+\dfrac{3}{4}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{4}\\x=-\dfrac{5}{4}\end{matrix}\right.\)

f: \(\left|4x\right|-\left|-13,5\right|=\left|2\dfrac{1}{4}\right|\)

=>\(4\left|x\right|=2,25+13,5=15,75\)

=>\(\left|x\right|=\dfrac{63}{16}\)

=>\(x=\pm\dfrac{63}{16}\)

g: \(\dfrac{5}{6}-\left|2-x\right|=\dfrac{1}{3}\)

=>\(\dfrac{5}{6}-\left|x-2\right|=\dfrac{1}{3}\)

=>\(\left|x-2\right|=\dfrac{5}{6}-\dfrac{1}{3}=\dfrac{1}{2}\)

=>\(\left[{}\begin{matrix}x-2=\dfrac{1}{2}\\x-2=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\)

h: \(\left|x-\dfrac{2}{5}\right|+\dfrac{1}{2}=\dfrac{3}{4}\)

=>\(\left|x-\dfrac{2}{5}\right|=\dfrac{3}{4}-\dfrac{1}{2}=\dfrac{1}{4}\)

=>\(\left[{}\begin{matrix}x-\dfrac{2}{5}=\dfrac{1}{4}\\x-\dfrac{2}{5}=-\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}+\dfrac{2}{5}=\dfrac{13}{20}\\x=-\dfrac{1}{4}+\dfrac{2}{5}=\dfrac{-5+8}{20}=\dfrac{3}{20}\end{matrix}\right.\)

i: \(\left|5-3x\right|+\dfrac{2}{3}=\dfrac{1}{6}\)

=>\(\left|3x-5\right|=\dfrac{1}{6}-\dfrac{2}{3}=\dfrac{1}{6}-\dfrac{4}{6}=-\dfrac{3}{6}=-\dfrac{1}{2}< 0\)

=>\(x\in\varnothing\)

k: \(-2,5+\left|3x+5\right|=-1,5\)

=>|3x+5|=-1,5+2,5=1

=>\(\left[{}\begin{matrix}3x+5=1\\3x+5=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-4\\3x=-6\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=-2\end{matrix}\right.\)

m: \(\dfrac{1}{5}-\left|\dfrac{1}{5}-x\right|=\dfrac{1}{5}\)

=>\(\left|\dfrac{1}{5}-x\right|=\dfrac{1}{5}-\dfrac{1}{5}=0\)

=>\(\dfrac{1}{5}-x=0\)

=>\(x=\dfrac{1}{5}\)

n: \(-\dfrac{22}{15}x+\dfrac{1}{3}=\left|-\dfrac{2}{3}+\dfrac{1}{5}\right|\)

=>\(-\dfrac{22}{15}x+\dfrac{1}{3}=\dfrac{2}{3}-\dfrac{1}{5}\)

=>\(-\dfrac{22}{15}x=\dfrac{1}{3}-\dfrac{1}{5}=\dfrac{2}{15}\)

=>-22x=2

=>\(x=-\dfrac{1}{11}\)

Lời giải:
Thể tích của bể: 

$4\times 3\times 0,5=6$ (m³)

Đổi $6$ m³ = $6000$ lít.

Mỗi phút cả 2 vòi chảy được: $85+25=110$ (lít)

Nếu cả hai vòi cùng chảy thì sẽ đầy bể sau:

$6000:110=\frac{600}{11}$ (phút)

\(\frac{1}{4}\) và \(\frac{1}{2}\)

Vì 0^x = 0 (Với mọi \(x\in R\)); 1^2x = 1 (Với mọi \(x\in Z\)).

Kẻ AH⊥BC

ta có: \(VP=AB^2+BC^2-2.AB.BC.cosB=AB^2+BC^2-2.AB.BC.\dfrac{BH}{AB}=AB^2+BC^2-2.BH.BC=AB^2-BH^2+BC^2-2.BH.BC+BH^2=AH^2+\left(BC-BH\right)^2=AH^2+CH^2=AC^2=VT\)

1) -2/2.3+(-2/3.4)+(-2/4.5)+...+(-2/19.20)

=-1(1/2-1/3+1/3-1/4+...+1/19-1/20)

=-1(1/2-1/20)

=-1.9/20

=-9/20

à nhầm

1)=-2(1/2-1/3+1/3-1/2+...+1/19-1/20)

=-2.(1/2-1/20)

=-2.9/20

=-9/10

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