giúp mình câu 84 và câu 87 ạ
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Ta có :
\(\frac{x-3}{97}+\frac{x-27}{73}+\frac{x-67}{33}+\frac{x-73}{27}=4\)
\(\Leftrightarrow\left(\frac{x-3}{97}-1\right)+\left(\frac{x-27}{73}-1\right)+\left(\frac{x-67}{33}-1\right)+\left(\frac{x-73}{27}-1\right)=0\)
\(\Leftrightarrow\frac{x-100}{97}+\frac{x-100}{73}+\frac{x-100}{33}+\frac{x-100}{27}=0\)
\(\Leftrightarrow\left(x-100\right)\left(\frac{1}{97}+\frac{1}{73}+\frac{1}{33}+\frac{1}{27}\right)=0\)
Vì \(\frac{1}{97}+\frac{1}{73}+\frac{1}{33}+\frac{1}{27}>0\) Nên \(x-100=0\)
\(\Leftrightarrow x=100\)
Vậy \(x=100\)
\(\Leftrightarrow\frac{x-3}{87}+\frac{x-27}{79}+\frac{x-67}{33}+\frac{x-73}{27}-4=0\)
\(\Leftrightarrow\left(\frac{x-3}{97}-1\right)+\left(\frac{x-27}{73}-1\right)+\left(\frac{x-67}{33}-1\right)+\left(\frac{x-73}{27}-1\right)=0\)
\(\Leftrightarrow\left(\frac{x-3-97}{97}\right)+\left(\frac{x-27-73}{73}\right)+\left(\frac{x-67-33}{33}\right)+\left(\frac{x-73-27}{27}\right)=0\)
\(\Leftrightarrow\frac{x-100}{97}+\frac{x-100}{73}+\frac{x-100}{33}+\frac{x-100}{27}=0\)
\(\Leftrightarrow\left(x-100\right)\left(\frac{1}{97}+\frac{1}{73}+\frac{1}{33}+\frac{1}{27}\right)=0\)
Vì \(\frac{1}{97}+\frac{1}{73}+\frac{1}{33}+\frac{1}{27}\ne0\)
\(\Rightarrow x-100=0\Leftrightarrow x=100\)
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d. \(\dfrac{\pi}{2}< a;b< \pi\Rightarrow sina>0;sinb>0\)
\(sina=\sqrt{1-cos^2a}=\dfrac{4}{5}\Rightarrow tana=\dfrac{sina}{cosa}=-\dfrac{4}{3}\)
\(sinb=\sqrt{1-cos^2b}=\dfrac{5}{13}\Rightarrow tanb=-\dfrac{5}{12}\)
Vậy:
\(sin\left(a-b\right)=sina.cosb-cosa.sinb=\dfrac{4}{5}.\left(-\dfrac{12}{13}\right)-\left(-\dfrac{3}{5}\right)\left(\dfrac{5}{13}\right)=...\)
\(cos\left(a-b\right)=cosa.cosb-sina.sinb=...\) (bạn tự thay số bấm máy)
\(tan\left(a+b\right)=\dfrac{tana+tanb}{1-tana.tanb}=...\)
\(cot\left(a+b\right)=\dfrac{1}{tan\left(a+b\right)}=\dfrac{1-tana.tanb}{tana+tanb}=...\)
e.
\(0< y< \dfrac{\pi}{2}\Rightarrow cosy>0\Rightarrow cosy=\sqrt{1-sin^2y}=\dfrac{4}{5}\)
\(\Rightarrow tany=\dfrac{siny}{cosy}=\dfrac{3}{4}\)
Vậy: \(tan\left(x+y\right)=\dfrac{tanx+tany}{1-tanx.tany}=...\)
\(cot\left(x-y\right)=\dfrac{1}{tan\left(x-y\right)}=\dfrac{1+tanx.tany}{tanx-tany}=...\)
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4 If you aren't alert, you computer will be infected with viruses
5 If we use much pesticide on vegetables, they will become poisonous
9 Because Nam needs to improve his social skills, he attends the activity
11 How about using public buses instead of motorbike
12 They enjoy protecting the environment unpolluted
15 Mr Brown was disappointed that his son didn't write to him
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Bài 11:
a: Ta có: \(P=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}\)
\(=\sqrt{x}\cdot\left(\sqrt{x}-1\right)\)
\(=x-\sqrt{x}\)
b: Để P=2 thì \(x-\sqrt{x}-2=0\)
hay x=4
Bài 10:
a: Ta có: \(A=\left(1+\dfrac{\sqrt{x}}{x+1}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}-x-1}\right)\)
\(=\dfrac{x+\sqrt{x}+1}{x+1}:\dfrac{x-2\sqrt{x}+1}{\left(x+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x+\sqrt{x}+1}{x+1}\cdot\dfrac{x+1}{\sqrt{x}-1}\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}-1}\)
b: Để A<0 thì \(\sqrt{x}-1< 0\)
hay x<1
Kết hợp ĐKXĐ, ta được: \(0\le x< 1\)
Để A=-1 thì \(x+\sqrt{x}+1=-\sqrt{x}+1\)
\(\Leftrightarrow x=0\)
c: Thay x=4 vào A, ta được:
\(A=\dfrac{4+2+1}{2-1}=7\)
84.
Ta có: \(\left\{{}\begin{matrix}SA\perp\left(ABC\right)\Rightarrow SA\perp BC\\BC\perp AB\left(gt\right)\end{matrix}\right.\) \(\Rightarrow BC\perp\left(SAB\right)\Rightarrow BC\perp SB\) đồng thời SB là hình chiếu vuông góc của SC lên (SAB)
\(\Rightarrow\widehat{BSC}\) là góc giữa SC và (SAB)
\(\Rightarrow\widehat{BSC}=30^0\)
\(\Rightarrow SB=\dfrac{BC}{tan30^0}=a\sqrt{3}\)
\(\Rightarrow SA=\sqrt{SB^2-AB^2}=a\sqrt{2}\)
\(V=\dfrac{1}{3}SA.BC^2=\dfrac{a^3\sqrt{2}}{3}\)
87.
Gọi M là trung điểm BC \(\Rightarrow AM\perp BC\) (trung tuyến đồng thời là đường cao trong tam giác cân)
Mà \(SA\perp\left(ABC\right)\Rightarrow SA\perp BC\)
\(\Rightarrow BC\perp\left(SAM\right)\)
Lại có BC là giao tuyến (SBC) và (ABC)
\(\Rightarrow\widehat{SMA}\) là góc giữa (SBC) và (ABC)
\(\Rightarrow\widehat{SMA}=60^0\)
\(AM=\sqrt{AB^2-BM^2}=\sqrt{4a^2-\left(\dfrac{3a}{2}\right)^2}=\dfrac{a\sqrt{7}}{2}\)
\(SA=AM.tan60^0=\dfrac{a\sqrt{21}}{2}\)
\(V=\dfrac{1}{3}SA.\dfrac{1}{2}.AM.BC=\dfrac{7a^3\sqrt{3}}{8}\)