HOC24
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= -7
\(\sqrt{A}\ge0\) ; \(\sqrt{B}\ge0\) Mà \(\sqrt{A}+\sqrt{B}=0\) => \(A=0;B=0\)
x = -1;0;1
* Theo định lí 1 ta có :
\(AB^2=BH.BC\) => \(BH=\dfrac{AB^2}{BC}\) <=> \(BH=\dfrac{4^2}{6}\)= \(\dfrac{8}{3}\)
\(\cos B=\dfrac{BH}{AB}=\dfrac{\dfrac{8}{3}}{4}=\dfrac{2}{3}\)
ấn shift cos \(\dfrac{2}{3}\approx48^0\)
Vậy góc B \(\approx48^0\)
* Theo định lí py-ta-go ta có :
\(BH^2+AH^2=AB^2\) => \(AH^2=AB^2-BH^2\) <=> \(AH^2=4^2-\left(\dfrac{8}{3}\right)^2=\dfrac{80}{9}\) => AH = \(\sqrt{\dfrac{80}{9}}=\dfrac{4\sqrt{5}}{3}\)
\(\cos CAH=\dfrac{AH}{AC}=\dfrac{\dfrac{4\sqrt{5}}{3}}{5}=\dfrac{4\sqrt{5}}{15}\)
ấn shift cos \(\dfrac{4\sqrt{5}}{15}\)\(\approx53^0\)
vậy góc C \(\approx53^0\)
góc A + góc B + góc C = 180 độ => góc A = 180 độ - ( góc B + góc C ) <=> góc A = 180 - ( 53 + 48 ) = 79 độ
Vậy góc A = \(79^0\)
3.215689741+3.25489= 3. ( 215689741 + 25489 ) = 3 . 215715230 = 647145690
góc \(A_1=70^0\) => góc BAC = \(180^o-70^o=110^o\)
mà góc C = 110 độ => góc C = góc BAC = \(110^o\)
=> 2 đường thẳng Ba và Cb // với nhau
mà góc D = \(90^0\) => B = \(90^0\)
=>BD vuông góc với AB
A= (\(\dfrac{\left(a\sqrt{a}-1\right)\cdot\left(a+\sqrt{a}\right)-\left(a\sqrt{a}+1\right)\cdot\left(a-\sqrt{a}\right)}{\left(a-\sqrt{a}\right)\cdot\left(a+\sqrt{a}\right)}\) )\(\cdot\left(\dfrac{\left(\sqrt{a}+1\right)\cdot\left(\sqrt{a}+1\right)+\left(\sqrt{a}-1\right)\cdot\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)\cdot\left(\sqrt{a}+1\right)}\right)\)
A = \(\left(\dfrac{a^2\sqrt{a}+a\sqrt{a^2}-a-\sqrt{a}-a^2\sqrt{a}-a\sqrt{a^2}+a+\sqrt{a}}{a^2-\sqrt{a^2}}\right)\) \(\cdot\left[\dfrac{\left(\sqrt{a}+1\right)^2+\left(\sqrt{a}-1\right)^2}{\sqrt{a^2}-1^2}\right]\)
A = \(\left(\dfrac{2a\sqrt{a^2}-2a}{a^2-\sqrt{a^2}}\right)\cdot\left[\dfrac{\left(\sqrt{a}+1\right)^2+\left(\sqrt{a}-1\right)^2}{a-1}\right]\)
A = \(\left[\dfrac{2\left(a^2-a\right)}{a^2-a}\right]\cdot\left[\dfrac{\left(\sqrt{a}+1\right)^2+\left(\sqrt{a}-1\right)^2}{a-1}\right]\)
A =\(2\cdot\left(\sqrt{a}+1\right)^2\cdot\left(\sqrt{a}-1\right)\)
α \(\approx\) 15,8 độ