ai giúp mình bài 2 vs =))))
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6.
Do \(AA'\perp\left(ABCD\right)\) (t/c hình hộp chữ nhật)
Mà \(AA'\in\left(ACC'A'\right)\)
\(\Rightarrow\left(ACC'A'\right)\perp\left(ABCD\right)\Rightarrow\) góc giữa (ACC'A') avf (ABCD) bằng 90 độ
b.
Từ H kẻ AH vuông góc BD (H thuộc BD)
Do \(AA'\perp\left(ABCD\right)\Rightarrow AA'\perp BD\)
\(\Rightarrow BD\perp\left(A'AH\right)\)
\(\Rightarrow\left\{{}\begin{matrix}BD\perp AH\\BD\perp A'H\end{matrix}\right.\)
Mà \(BD=\left(A'BD\right)\cap\left(ABCD\right)\Rightarrow\widehat{AHA'}\) là góc giữa (A'BD) và (ABCD)
\(AH=\dfrac{AB.AD}{\sqrt{AB^2+AD^2}}=\dfrac{bc}{\sqrt{b^2+c^2}}\)
\(\Rightarrow tan\widehat{AHA'}=\dfrac{AA'}{AH}=\dfrac{a\sqrt{b^2+c^2}}{bc}\)
7.
Kẻ \(AI\perp CM\Rightarrow\widehat{IAM}=\widehat{BCM}\) (góc có cạnh tương ứng vuông góc)
\(CM=\sqrt{BC^2+BM^2}=\sqrt{BC^2+\left(\dfrac{AB}{2}\right)^2}=2a\)
\(\Rightarrow AI=AM.cos\widehat{IAM}=\dfrac{AB}{2}.cos\widehat{BCM}=\dfrac{AB}{2}.\dfrac{BC}{CM}=\dfrac{a\sqrt{3}}{2}\)
b.
\(\left\{{}\begin{matrix}SA\perp\left(ABC\right)\Rightarrow SA\perp CI\\CI\perp AI\left(gt\right)\end{matrix}\right.\) \(\Rightarrow CI\perp\left(SAI\right)\Rightarrow\left\{{}\begin{matrix}CI\perp SI\\CI\perp AI\end{matrix}\right.\)
Mà \(CI=\left(SMC\right)\cap\left(ABC\right)\Rightarrow\widehat{SIA}\) là góc giữa (SMC) và (ABC)
\(tan\widehat{SIA}=\dfrac{SA}{AI}=\dfrac{4\sqrt{3}}{3}\Rightarrow\widehat{SIA}\approx66^035'\)
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Lời giải:
ĐKXĐ: $x\neq \pm 3; x\neq 0$
a. \(A=\left[\frac{-(x-3)}{x+3}.\frac{(x+3)^2}{(x-3)(x+3)}+\frac{x}{x+3}\right].\frac{x+3}{3x^2}\)
\(=\left(-1+\frac{x}{x+3}\right).\frac{x+3}{3x^2}=\frac{-3}{x+3}.\frac{x+3}{3x^2}=\frac{-1}{x^2}\)
b. Với $x=\frac{-1}{2}$ thì $x^2=\frac{1}{4}$
$\Rightarrow A=\frac{-1}{\frac{1}{4}}=-4$
c.
Với $x\neq 0, \pm 3$ thì $\frac{1}{x^2}>0\Leftrightarrow A=\frac{-1}{x^2}< 0$ với mọi $x\neq 0; x\neq \pm 3$
a) Ta có: \(A=\left(\dfrac{3-x}{x+3}\cdot\dfrac{x^2+6x+9}{x^2-9}+\dfrac{x}{x+3}\right):\dfrac{3x^2}{x+3}\)
\(=\left(\dfrac{-\left(x-3\right)}{x+3}\cdot\dfrac{\left(x+3\right)^2}{\left(x+3\right)\left(x-3\right)}+\dfrac{x}{x+3}\right):\dfrac{3x^2}{x+3}\)
\(=\left(\dfrac{-x-3+x}{x+3}\right)\cdot\dfrac{x+3}{3x^2}\)
\(=-\dfrac{1}{x^2}\)
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4a C IV ; S II ; NO3 I ;Br I b) K I ; Ba II ;S IV 5a Fe II b Fe III
2:
a: |2x-3|=7
=>2x-3=7 hoặc 2x-3=-7
=>x=5(loại) hoặc x=-2(nhận)
Khi x=-2 thì \(B=\dfrac{-2+1}{2\cdot\left(-2\right)^2-10\cdot\left(-2\right)}=\dfrac{-1}{2\cdot4+20}=\dfrac{-1}{28}\)
b: \(A=\dfrac{15-x+2x-10}{\left(x-5\right)\left(x+5\right)}=\dfrac{1}{x-5}\)
c: P=A:B
\(=\dfrac{1}{x-5}\cdot\dfrac{2x\left(x-5\right)}{x+1}=\dfrac{2x}{x+1}\)
P>2
=>P-2>0
=>(2x-x-1)/(x+1)>0
=>(x-1)/(x+1)>0
=>x>1 hoặc x<-1