a) Tình diện tích hình có gạch chéo
b) So sánh diện tích hình có gạch chéo và diện tích hình tam giác IBC
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+ Ta có:
3√3+1=3(√3−1)(√3+1)(√3−1)=3√3−3.1(√3)2−1233+1=3(3−1)(3+1)(3−1)=33−3.1(3)2−12
=3√3−33−1=3√3−32=33−33−1=33−32.
+ Ta có:
2√3−1=2(√3+1)(√3−1)(√3+1)=2(√3+1)(√3)2−1223−1=2(3+1)(3−1)(3+1)=2(3+1)(3)2−12
=2(√3+1)3−1=2(√3+1)2=√3+1=2(3+1)3−1=2(3+1)2=3+1.
+ Ta có:
2+√32−√3=(2+√3).(2+√3)(2−√3)(2+√3)=(2+√3)222−(√3)22+32−3=(2+3).(2+3)(2−3)(2+3)=(2+3)222−(3)2
=22+2.2.√3+(√3)24−3=22+2.2.3+(3)24−3=4+4√3+31=(4+3)+4√31=4+43+31=(4+3)+431
=7+4√31=7+4√3=7+431=7+43.
+ Ta có:
b3+√b=b(3−√b)(3+√b)(3−√b)b3+b=b(3−b)(3+b)(3−b)
=b(3−√b)32−(√b)2=b(3−√b)9−b;(b≠9)=b(3−b)32−(b)2=b(3−b)9−b;(b≠9).
+ Ta có:
p2√p−1=p(2√p+1)(2√p−1)(2√p+1)p2p−1=p(2p+1)(2p−1)(2p+1)
=p(2√p+1)(2√p)2−12=p(2√p+1)4p−1=p(2p+1)(2p)2−12=p(2p+1)4p−1=2p√p+p4p−1
Bài 51 trang 30 SGK Toán 9 tập 1 - loigiaihay.com
#Ye Chi-Lien
\(\frac{3}{\sqrt{3}+1}=\frac{3\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\frac{3\sqrt{3}-3}{3-1}=\frac{3\sqrt{3}-3}{2}\)
\(\frac{2}{\sqrt{3}-1}=\frac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}=\frac{2\left(\sqrt{3}+1\right)}{3-1}=\sqrt{3}-1\)
\(\frac{2+\sqrt{3}}{2-\sqrt{3}}=\frac{\left(2+\sqrt{3}\right)^2}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=4-3}=\left(2+\sqrt{3}\right)^2=4+4\sqrt{3}+3=7+4\sqrt{3}\)
\(\frac{b}{3+\sqrt{b}}=\frac{b\left(3-\sqrt{b}\right)}{\left(3+\sqrt{b}\right)\left(3-\sqrt{b}\right)}=\frac{b\left(3-\sqrt{b}\right)}{9-b}\)
\(\frac{p}{2\sqrt{p}-1}=\frac{p\left(2\sqrt{p}+1\right)}{\left(2\sqrt{p}-1\right)\left(2\sqrt{b}+1\right)}=\frac{p\left(2\sqrt{b}+1\right)}{4p-1}\)
\(\frac{5}{\sqrt{10}}=\frac{5\sqrt{10}}{10}=\frac{\sqrt{10}}{2}\)
\(\frac{5}{2\sqrt{5}}=\frac{10\sqrt{5}}{20}=\frac{\sqrt{5}}{2}\)
\(\frac{1}{3\sqrt{20}}=\frac{3\sqrt{20}}{180}=\frac{\sqrt{20}}{60}=\frac{2\sqrt{5}}{60}=\frac{\sqrt{5}}{30}\)
\(\frac{2\sqrt{2}+2}{5\sqrt{2}}=\frac{10\sqrt{2}\left(\sqrt{2}+1\right)}{50}=\frac{20+10\sqrt{2}}{50}=\frac{10\left(2+\sqrt{2}\right)}{50}=\frac{2+\sqrt{2}}{5}\)
\(\frac{y+b\sqrt{y}}{b\sqrt{y}}=\frac{y\left(\sqrt{y}+b\right)}{by}=\frac{\sqrt{y}+b}{b}\)
+ Ta có:
5√10=5.√10√10.√10=5√10(√10)2=5√1010510=5.1010.10=510(10)2=51010
=5.√105.2=5.105.2=√102=102.
+ Ta có:
52√5=5.√52√5.√5=5√52.(√5.√5)=5√52(√5)2525=5.525.5=552.(5.5)=552(5)2
=5√52.5=√52=552.5=52.
+ Ta có:
13√20=1.√203√20.√20=√203.(√20.√20)=√203.(√20)21320=1.20320.20=203.(20.20)=203.(20)2
=√203.20=√22.560=2√560=2√52.30=√530=203.20=22.560=2560=252.30=530.
+ Ta có:
(2√2+2)5.√2=(2√2+2).√25√2.√2=2√2.√2+2.√25.(√2)2(22+2)5.2=(22+2).252.2=22.2+2.25.(2)2
=2.2+2√25.2=2(2+√2)5.2=2+√25=2.2+225.2=2(2+2)5.2=2+25.
+ Ta có:
y+b√yb√y=(y+b√y).√yb√y.√y=y√y+b√y.√yb.(√y)2y+byby=(y+by).yby.y=yy+by.yb.(y)2
=y√y+b(√y)2by=y√y+byby=yy+b(y)2by=yy+byby
=y(√y+b)b.y=√y+bb=y(y+b)b.y=y+bb.
Cách khác:
y+b√yb√y=(√y)2+b√yb√yy+byby=(y)2+byby=√y(√y+b)b√y=√y+bb
Nguồn : Bài 50 trang 30 SGK Toán 9 tập 1 - loigiaihay.com
#Ye Chi-Lien
(do xy > 0 (gt) nên đưa thừa số xy vào trong căn để khử mẫu)
#Học tốt!!!
\(ab\cdot\sqrt{\dfrac{a}{b}}=a\cdot\sqrt{ab}\)
\(\dfrac{a}{b}\cdot\sqrt{\dfrac{b}{a}}=\dfrac{\sqrt{a\cdot b}}{b}\)
\(\sqrt{\dfrac{1}{b}+\dfrac{1}{b^2}}=\dfrac{\sqrt{b+1}}{b}\)
\(\sqrt{\dfrac{9\cdot a^3}{36\cdot b}}=\dfrac{\sqrt{a^3\cdot b}}{2\cdot b}\)
\(3\cdot x\cdot y\cdot\sqrt{\dfrac{2}{x\cdot y}}=3\cdot\sqrt{2\cdot x\cdot y}\)
\(\sqrt{\dfrac{1}{600}}\)=\(\sqrt{\dfrac{1}{10^2\cdot6}}\)=\(\sqrt{\dfrac{1\cdot6}{10^2\cdot6\cdot6}}\)=\(\dfrac{\sqrt{6}}{60}\)
\(\sqrt{\dfrac{11}{540}}\)=\(\sqrt{\dfrac{11\cdot540}{540\cdot540}}\)=\(\dfrac{\sqrt{5940}}{540}\)=\(\dfrac{\sqrt{165}}{90}\)
\(\sqrt{\dfrac{3}{50}}\)=\(\sqrt{\dfrac{3\cdot50}{50\cdot50}}\)=\(\dfrac{\sqrt{150}}{50}\)=\(\dfrac{\sqrt{6}}{10}\)
\(\sqrt{\dfrac{5}{98}}\)=\(\sqrt{\dfrac{5\cdot98}{98\cdot98}}=\dfrac{\sqrt{490}}{98}=\dfrac{\sqrt{10}}{14}\)
\(\sqrt{\dfrac{\left(1-\sqrt{3}\right)^2}{27}}=\dfrac{3-\sqrt{3}}{9}\)
\(\sqrt{\dfrac{1}{600}}=\dfrac{\sqrt{6}}{60}\)
\(\sqrt{\dfrac{11}{540}}=\dfrac{\sqrt{165}}{90}\)
\(\sqrt{\dfrac{3}{50}}=\dfrac{\sqrt{6}}{10}\)
\(\sqrt{\dfrac{5}{98}}=\dfrac{\sqrt{10}}{14}\)
\(\sqrt{\dfrac{\left(1-\sqrt{3}\right)^2}{27}}=\dfrac{3-\sqrt{3}}{9}\)
Ngày thứ 3 cửa hàng còn đc số gạo là:
1-3838 -50%=1818 (số gạo)
Số gạo của cửa hàng đó là:
10×8=80(tấn)
Đ/S:80 tấn