Nguyễn Thiên Phúc
Giới thiệu về bản thân
\(\bullet\)Có \(\triangle\)ACD vuông tại A(AC\(\bot\)BA)
Nên tan \(\widehat{ACD}\)=\(\dfrac{AD}{AC}\)(TSLG của góc nhọn)
tan \(\widehat{ACD}\)=\(\dfrac{20}{30}\)=\(\dfrac{2}{3}\)
\(\widehat{ACD}\approx\)33\(^\circ\)
\(\bullet\)Có CD\(\bot\)phương BC(GT)
Nên \(\widehat{BCD}\)=90\(^\circ\)
\(\widehat{BCA}\)+\(\widehat{ACD}\)=\(\widehat{BCD}\)
\(\widehat{ACB}\) =\(\widehat{BCD}\)-\(\widehat{ACD}\)
\(\widehat{ACB}\) \(\approx\) 90\(^\circ\) - 33\(^\circ\)
\(\widehat{ACB}\) \(\approx\) 57\(^\circ\)
\(\bullet\)Có \(\triangle\)BCA vuông tại A(AC\(\bot\)BA)
Nên tan \(\widehat{BCA}\)=\(\dfrac{BA}{AC}\)(TSLG của góc nhọn)
tan 57\(^\circ\) \(\approx\)\(\dfrac{AB}{30}\)
tan 57\(^\circ\bullet\)30\(\approx\)AB
46 \(\approx\)AB(m)
a) ĐKXĐ: a≠0, 1
P=\(\left(\dfrac{a}{2\sqrt{a}}-\dfrac{1}{2\sqrt{a}}\right)^2\bullet\left(\dfrac{\left(\sqrt{a}-1\right)^2}{a-1}-\dfrac{\left(\sqrt{a}+1\right)^2}{a-1}\right)\)
P=\(\dfrac{\left(a-1\right)^2}{4a}\bullet\dfrac{\left(a-2\sqrt{a}+1\right)-\left(a+2\sqrt{a}+1\right)}{a-1}\)
P=\(\dfrac{a-1}{4a}\bullet\)(a-2\(\sqrt{a}\)+1-a-2\(\sqrt{a}\)-1)
P=\(\dfrac{a-1}{4a}\bullet\)[(a-a)+(-2\(\sqrt{a}\)-2\(\sqrt{a}\))+(1-1)]
P=\(\dfrac{1-a}{\sqrt{a}}\)
b) ĐKXĐ: a≠0, 1
P<0
\(\dfrac{1-a}{\sqrt{a}}\)<0
a>1(TM)
Vậy a>1 để P<0
a) ĐKXĐ: a≠0, 1
P=\(\left(\dfrac{a}{2\sqrt{a}}-\dfrac{1}{2\sqrt{a}}\right)^2\bullet\left(\dfrac{\left(\sqrt{a}-1\right)^2}{a-1}-\dfrac{\left(\sqrt{a}+1\right)^2}{a-1}\right)\)
P=\(\dfrac{\left(a-1\right)^2}{4a}\bullet\dfrac{\left(a-2\sqrt{a}+1\right)-\left(a+2\sqrt{a}+1\right)}{a-1}\)
P=\(\dfrac{a-1}{4a}\bullet\)(a-2\(\sqrt{a}\)+1-a-2\(\sqrt{a}\)-1)
P=\(\dfrac{a-1}{4a}\bullet\)[(a-a)+(-2\(\sqrt{a}\)-2\(\sqrt{a}\))+(1-1)]
P=\(\dfrac{1-a}{\sqrt{a}}\)
b) ĐKXĐ: a≠0, 1
P<0
\(\dfrac{1-a}{\sqrt{a}}\)<0
a>1(TM)
Vậy a>1 để P<0
A=4\(\sqrt{3}\)-4\(\sqrt{3}\)+9\(\sqrt{3}\)-6\(\sqrt{3}\)
A= 3\(\sqrt{3}\)
A=4\(\sqrt{3}\)-4\(\sqrt{3}\)+9\(\sqrt{3}\)-6\(\sqrt{3}\)
A= 3\(\sqrt{3}\)
a) 3x-4=5+x b) 3(x-1)-7=5(x+2)
3x-x=9 3x-3 -7= 5x+10
2x=9 3x-5x=20
x=4,5 -2x =20
Vầy có nghiệm x=4,5 x =-10
Vầy có nghiệm x=-10
a) 3x-5=4 b) \(\dfrac{2x}{3}\)+\(\dfrac{3x-1}{6}\)=\(\dfrac{x}{2}\)
3x=9 \(\dfrac{4x}{6}\)+\(\dfrac{3x-1}{6}\)=\(\dfrac{3x}{6}\)
x=3 4x+3x-1=3x
Vầy x=3 7x-3x=1
4x=1
x=0,25
Vầy x=0,25