\(\dfrac{1}{7}\left(\dfrac{7}{3.10}+\dfrac{7}{10.17}+...+\dfrac{7}{73.80}-\left(\dfrac{7}{2.9}+\dfrac{7}{9.16}+...+\dfrac{7}{23.30}\right)\right)\)

\(=\dfrac{1}{7}\left(\dfrac{1}{3}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{17}+...+\dfrac{1}{73}-\dfrac{1}{80}-\left(\dfrac{1}{2}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{23}-\dfrac{1}{30}\right)\right)\)

\(=\dfrac{1}{7}\left(\dfrac{1}{3}-\dfrac{1}{80}-\left(\dfrac{1}{2}-\dfrac{1}{30}\right)\right)\)

\(=\dfrac{1}{7}\left(\dfrac{77}{240}-\dfrac{7}{15}\right)=\dfrac{1}{7}.\left(-\dfrac{7}{48}\right)=-\dfrac{1}{48}\)

\(53.74-53.7^2+5^2.47\)

\(=53.74-53.49+25.47\)

\(=53\left(74-49\right)+25.47\)

\(=53.25+25.47\)

\(=25\left(53+47\right)=25.100=2500\)

a) \(7^2-7\left(13-x\right)=14\)

\(7\left(13-x\right)=49-14=35\)

\(13-x=5\)

\(x=13-5=8\)

b) \(5x-5^2=10\)

\(5x=10+25=35\)

\(x=7\)

c) \(4\left(x-5\right)-2^3=2^4.3=48\)

\(4\left(x-5\right)=48+8=56\)

\(x-5=14\)

\(x=19\)

\(27+27+72:9\)

\(=54+8\)

\(=62\)

\(sinC=\dfrac{AH}{AC}=\dfrac{9}{16}\)

\(\Rightarrow\widehat{C}\simeq34,2\)

\(\Rightarrow\widehat{B}=180^o-90^o-34,2^o=55,8^o\)

\(\left\{{}\begin{matrix}sinB=\dfrac{AC}{BC}\\cosB=\dfrac{AB}{BC}\\tanB=\dfrac{AC}{AB}\\cotB=\dfrac{AB}{AC}\end{matrix}\right.\)

a) \(210:x=2,1:0,25=8,4\)

\(x=210:8,4=25\)

b) \(x.2+x.0,5=6,25\)

\(2,5x=6,25\)

\(x=6,25:2,5=2,5\)

c) \(x:0,25-x=15,6\)

\(3x=15,6\)

\(x=5,2\)

d) \(x:12,5=3,75.4=15\)

\(x=15.12,5=187,5\)

e) \(x.12,6-x.5,6=42\)

\(7x=42\)

\(x=6\)

g) \(x:0,1-x:4-x.0,75=2,25\)

\(10x-\dfrac{x}{4}-\dfrac{3x}{4}=2,25\)

\(9x=2,25\)

\(x=0,25\)

Ta có:

\(\left\{{}\begin{matrix}GM=\dfrac{1}{3}AM=5\\GN=\dfrac{1}{3}BN=4\end{matrix}\right.\)

\(S_{CMN}=S_{AMN}=15\sqrt{3}\)

\(S_{GMN}=\dfrac{1}{3}S_{AMN}=5\sqrt{3}\)

=> \(5\sqrt{3}=\dfrac{1}{2}.GM.GN.sinG\)

\(\Rightarrow sinG=\dfrac{\sqrt{3}}{2}\Rightarrow\widehat{G}=60^o\)

\(MN=\sqrt{GM^2+GN^2-2GM.GN.cosG}=\sqrt{21}\)

\(x=\sqrt{9+4\sqrt{5}}-\sqrt{6+2\sqrt{5}}\)

\(=\sqrt{\left(\sqrt{5}+2\right)^2}-\sqrt{\left(\sqrt{5}+1\right)^2}\)

\(=\sqrt{5}+2-\sqrt{5}-1=1\) ( \(\left\{{}\begin{matrix}\sqrt{5}+2>0\\\sqrt{5}+1>0\end{matrix}\right.\))

a) \(A=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}=\dfrac{1-1}{1+2}=0\left(đk:x\ge0\right)\)

b) \(A>\dfrac{1}{2}\)\(\Rightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}+2}>\dfrac{1}{2}\)

\(\Leftrightarrow\dfrac{2\sqrt{x}-2-\sqrt{x}-2}{2\sqrt{x}+4}>0\)

\(\Leftrightarrow\sqrt{x}>4\) (vì \(2\sqrt{x}+4>0\))

\(\Leftrightarrow x>16\)

a) \(x^2+4x+4-4y^2=\left(x+2\right)^2-\left(2y\right)^2=\left(x+2y+2\right)\left(x-2y+2\right)\)

b) \(x^2+6xy+9y^2-4z^2=\left(x+3y\right)^2-\left(2z\right)^2=\left(x+3y-2z\right)\left(x+3y+2z\right)\)

c) \(x^2+2xy+y^2-z^2+4zt-4t^2=\left(x+y\right)^2-\left(z-2t\right)^2\)

\(=\left(x+y+z-2t\right)\left(x+y-z+2t\right)\)

d) \(a^2-x^2-ay+xy=\left(a-x\right)\left(a+x\right)-y\left(a-x\right)=\left(a-x\right)\left(a+x-y\right)\)

e) \(x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)\)

\(=x^2\left(y-z\right)-y^2\left(y-z+x-y\right)+z^2\left(x-y\right)\)

\(=\left(x^2-y^2\right)\left(y-z\right)+\left(z^2-y^2\right)\left(x-y\right)\)

\(=\left(y-z\right)\left(x^2-y^2-\left(y+z\right)\left(x-y\right)\right)\)

\(=\left(y-z\right)\left(x^2-y^2-xy+y^2-zx+yz\right)\)

\(=\left(x-y\right)\left(y-z\right)\left(z-x\right)\)

a) \(x^4+3x^3+x^2+3x=x^3\left(x+3\right)+x\left(x+3\right)=x\left(x+3\right)\left(x^2+1\right)\)

b) \(x^4+x^2-27x-9=x\left(x^3-27\right)+\left(x-3\right)\left(x+3\right)\)

\(=x\left(x-3\right)\left(x^2+3x+9\right)+\left(x-3\right)\left(x+3\right)\)

\(=\left(x-3\right)\left(x^3+3x^2+9x+x+3\right)=\left(x-3\right)\left(x^3+3x^2+10x+3\right)\)

c) \(x^2-xy-x+y=x\left(x-y\right)-\left(x-y\right)=\left(x-1\right)\left(x-y\right)\)

d) \(xy+4-x^2+2y=\left(2-x\right)\left(x+2\right)+y\left(x+2\right)=\left(x+2\right)\left(y+2-x\right)\)

e) \(xy+y-2\left(x+1\right)=y\left(x+1\right)-2\left(x+1\right)=\left(y-2\right)\left(x+1\right)\)

f) \(5\left(x-y\right)+ax-ay=5\left(x-y\right)+a\left(x-y\right)=\left(a+5\right)\left(x-y\right)\)