Giải chi tiết phần b bài 1 và phần b bài 2 giúp em với ạ
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1: góc DMB+góc DHB=180 độ
=>DMBH nội tiếp
2: Kẻ tiếp tuyến Ax của (O)
=>góc xAC=góc ABC
góc AMD+góc AND=180 độ
=>AMDN nội tiếp
=>góc ANM=góc ADM=góc ABH
=>góc ANM=góc xAC
=>Ax//MN
3:
b: x1^2+x2^2=12
=>(x1+x2)^2-2x1x2=12
=>(2m+2)^2-4m=12
=>4m^2+4m+4=12
=>m^2+m+1=3
=>(m+2)(m-1)=0
=>m=1;m=-2
2:
b: =>|x1|-|x2|=m+3-|-1|=m+2
=>x1^2+x2^2-2|x1x2|=m+2
=>(x1+x2)^2-2x1x2-2|x1x2|=m+2
=>(2m)^2-2(-1)-2|-1|=m+2
=>4m^2-m-2=0
=>m=(1+căn 33)/8; m=(1-căn 33)/8
\(2x=3y\\ =>\dfrac{x}{3}=\dfrac{y}{2}\\ 4y=5z\\ =>\dfrac{y}{5}=\dfrac{z}{4}\\ \dfrac{x}{3}=\dfrac{y}{2}\\ =>\dfrac{x}{3.5}=\dfrac{y}{2.5}\\ =>\dfrac{x}{15}=\dfrac{y}{10}\\ \dfrac{y}{5}=\dfrac{z}{4}\\ =>\dfrac{y}{5.2}=\dfrac{z}{4.2}\\ =>\dfrac{y}{10}=\dfrac{z}{8}\\ =>\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{8}\)
Áp dụng tính chất dãy tỉ số bằng nhau
\(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{8}=\dfrac{x+y+z}{15+10+8}=\dfrac{11}{33}=\dfrac{1}{3}\\ =>\left\{{}\begin{matrix}x=\dfrac{1}{3}.15=5\\y=\dfrac{1}{3}.10=\dfrac{10}{3}\\z=\dfrac{1}{3}.8=\dfrac{8}{3}\end{matrix}\right.\)
\(a,\dfrac{3^{10}.11+9^5.5}{27^3.2^4}.x=-9\\ =>\dfrac{3^{10}.11+\left(3^2\right)^5.5}{\left(3^3\right)^3.2^4}.x=-9\\ =>\dfrac{3^{10}.\left(11+5\right)}{3^9.2^4}.x=-9\\ =>\dfrac{3^{10}.16}{3^9.2^4}.x=-9\\ =>\dfrac{3^{10}.2^4}{3^9.2^4}.x=-9\\ =>3^1.x=-9\\ =>x=-9:3\\ =>x=-3\)
Giải
\(\sqrt{\dfrac{9}{4}}-\left|2x+1\right|=0,75\)
TH1: \(\left|2x+1\right|=2x+1\)
\(=>\sqrt{\dfrac{9}{4}}-\left(2x+1\right)=0,75\\ =>\dfrac{3}{2}-2x-1=\dfrac{3}{4}\\ =>\left(\dfrac{3}{2}-1\right)-2x=\dfrac{3}{4}\\ =>\dfrac{1}{2}-2x=\dfrac{3}{4}\\ =>2x=\dfrac{1}{2}-\dfrac{3}{4}\\ =>2x=-\dfrac{1}{4}\\ =>x=\left(-\dfrac{1}{4}\right):2\\ =>x=-\dfrac{1}{8}\)
\(TH2:\left|2x+1\right|=-2x-1\\ =>\sqrt{\dfrac{9}{4}}-\left(-2x-1\right)=\dfrac{3}{4}\\ =>\dfrac{3}{2}+2x+1=\dfrac{3}{4}\\ =>\left(\dfrac{3}{2}+1\right)+2x=\dfrac{3}{4}\\ =>\dfrac{5}{2}+2x=\dfrac{3}{4}\\ =>2x=\dfrac{3}{4}-\dfrac{5}{2}\\ =>2x=-\dfrac{7}{4}\\ =>x=\left(-\dfrac{7}{4}\right):2\\ =>x=-\dfrac{7}{8}\)
Giải
\(a,5-\left(x-2\right)^2=-4\\ =>\left(x-2\right)^2=5-\left(-4\right)\\ =>\left(x-2\right)^2=9\\ =>\left[{}\begin{matrix}\left(x-2\right)^2=3^2\\\left(x-2\right)^2=\left(-3\right)^2\end{matrix}\right.=>\left[{}\begin{matrix}x-2=3\\x-2=-3\end{matrix}\right.=>\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
Vậy \(x=5;x=-1\)
Bài 2:
b: Ta có: \(B=\dfrac{15-5\sqrt{x}}{x-5\sqrt{x}+6}+\dfrac{\sqrt{x}+3}{\sqrt{x}-2}\)
\(=\dfrac{-5\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}+3}{\sqrt{x}-2}\)
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}-2}=1\)