Cho tam giác ABC vuông tại A, có AB=12cm, AC=16cm. Kẻ đường cao AH(H thuộc BC).

Xét \(\Delta HBA\) và \(\Delta ABC\) có:

        \(\widehat{ABC}\)chung

 \(\widehat{BHA}=\widehat{BAC}\left(=90^o\right)\)

\(\Rightarrow\Delta HBA~\Delta ABC\left(g.g\right)\)

b.AD ĐL Pitago vào \(\Delta ABC\) vuông tại A có:

\(BC^2=AB^2+AC^2\)

\(BC^2=12^2+16^2\)

\(BC^2=144+256=400\)

\(BC=\sqrt{400}=20\left(cm\right)\)

Vì \(\Delta HBA~\Delta ABC\)

\(\Rightarrow\frac{AH}{AB}=\frac{AC}{BC}\)

\(\Rightarrow\frac{AH}{12}=\frac{16}{20}\Rightarrow AH=\frac{12.16}{20}=9,6\left(cm\right)\)

ko

mình lớp 5

ko em lớp 4

a) Cm tamgiac ABC đồng dạng với tamgiac HBA(g.g)

=> AB/BC = BH/AB hay AB^2 = BH.HC

và cm  tamgiac ABC đồng dạng với tamgiac HAC(g.g)

=> AC/BC = HC/AC hay AC^2 = CH.BH

a. Xét tg vuông ABC và  tg vuông HBA có:

\(\widehat{ABH}\)chung

\(\Rightarrow\Delta ABC~\Delta HBA\)

\(\Rightarrow\frac{AB}{HB}=\frac{BC}{BA}\)

\(\Rightarrow AB^2=HB.BC\)

Cmtt:\(\Delta ABC~HAC\)

\(\Rightarrow\frac{AC}{HC}=\frac{BC}{AC}\)

\(\Rightarrow AC^2=BC.HC\)

b. lát làm tiếp nhá

x+1/2016+x+3/2018+x+5/1010+x+7/1011=6

x+1/2016+x+3/2018+x+5/1010+x+7/1011 -6=0

x+1/2016 -1 +x+3/2018 -1 + x+5/1010 -2 +x+7/1011 -2=0

x+1-2016/2016 + x+3-2018/2018 + x+5-2020/1010 +x+7-2022/ 1011 =0

x-2015/2016 +x-2015/2018 + x-2015/1010 + x-2015/1011=0

x-2015(1/2016 +1/ 2018 +1/ 1010 + 1/1011)=0
mà 1/2016+1/2018+1/1010+1/1011>0

<=>x-2015=0

x=2015

\(\frac{x+1}{2016}+\frac{x+3}{2018}+\frac{x+5}{1010}+\frac{x+7}{1011}=6\)

\(\Leftrightarrow\left(\frac{x+1}{2016}-1\right)+\left(\frac{x+3}{2018}-1\right)+\left(\frac{x+5}{1010}-2\right)+\left(\frac{x+7}{1011}-2\right)=0\)

\(\Leftrightarrow\frac{x-2015}{2016}+\frac{x-2015}{2018}+\frac{x-2015}{1010}+\frac{x-2015}{1011}=0\)

\(\Leftrightarrow\left(x-2015\right)\left(\frac{1}{2016}+\frac{1}{2018}+\frac{1}{1010}+\frac{1}{1011}\right)=0\)
\(\Leftrightarrow x-2015=0\)

\(\Leftrightarrow x=2015\)

a.\(2x\left(7x^2-5x-1\right)=14x^3-10x^2-2x\)

b.\(-2x^3\left(2x^2-3y+5yz\right)=-4x^5+6x^3y-10x^3yz\)

c.\(\left(2x-y\right)\left(4x^2-2xy+y^2\right)=2x\left(4x^2-2xy+y^2\right)-y\left(4x^2-2xy+y^2\right)\)

\(=8x^2-4x^2y+4xy^2-4x^2y+2xy^2-y^3\)

a.2x(7x2−5x−1)=14x3−10x2−2x

b.−2x3(2x2−3y+5yz)=−4x5+6x3y−10x3yz

c.(2x−y)(4x2−2xy+y2)=2x(4x2−2xy+y2)−y(4x2−2xy+y2)

=8x2−4x2y+4xy2−4x2y+2xy2−y3

Đặt \(A=\sqrt{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\left(\frac{\sqrt{a+b}}{c}+\frac{\sqrt{b+c}}{a}+\frac{\sqrt{c+a}}{b}\right)\)\(\left(a,b,c>0\right)\).

\(A=\frac{\left(a+b\right)\sqrt{\left(b+c\right)\left(c+a\right)}}{c}+\frac{\left(b+c\right)\sqrt{\left(c+a\right)\left(a+b\right)}}{a}\)\(+\frac{\left(c+a\right)\sqrt{\left(a+b\right)\left(b+c\right)}}{b}\).

\(A=\left(a+b\right)\sqrt{\frac{bc+ab+c^2+ac}{c^2}}+\left(b+c\right)\sqrt{\frac{ac+bc+a^2+ab}{a^2}}\)\(+\left(c+a\right)\sqrt{\frac{ab+ac+b^2+bc}{b^2}}\).

\(A=\left(a+b\right)\sqrt{\frac{a+b}{c}+\frac{ab}{c^2}+1}+\left(b+c\right)\sqrt{\frac{b+c}{a}+\frac{bc}{a^2}+1}\)\(+\left(c+a\right)\sqrt{\frac{c+a}{b}+\frac{ca}{b^2}+1}\)

Vì \(a,b,c>0\)nên áp dụng bất đẳng thức Cô-si cho 2 số dương, ta được:

\(a+b\ge2\sqrt{ab}\).

\(\Leftrightarrow\frac{a+b}{c}\ge\frac{2\sqrt{ab}}{c}\).

\(\Leftrightarrow\frac{a+b}{c}+1+\frac{ab}{c^2}\ge\frac{2\sqrt{ab}}{c}+1+\frac{ab}{c^2}=\left(\frac{\sqrt{ab}}{c}+1\right)^2\).

\(\Leftrightarrow\sqrt{\frac{a+b}{c}+1+\frac{ab}{c^2}}\ge\frac{\sqrt{ab}}{c}+1\).

\(\Leftrightarrow\left(a+b\right)\sqrt{\frac{a+b}{c}+1+\frac{ab}{c^2}}\ge\frac{\left(a+b\right)\sqrt{ab}}{c}+\left(a+b\right)\)\(\left(1\right)\).

Ta lại có \(\frac{a+b}{c}\ge\frac{2\sqrt{ab}}{c}\)(chứng minh trên).

\(\Leftrightarrow\frac{\left(a+b\right)\sqrt{ab}}{c}\ge\frac{2ab}{c}\).

\(\Leftrightarrow\frac{\left(a+b\right)\sqrt{ab}}{c}+\left(a+b\right)\ge\frac{2ab}{c}+\left(a+b\right)\)\(\left(2\right)\).

Từ \(\left(1\right)\)và \(\left(2\right)\), ta được:

\(\left(a+b\right)\sqrt{\frac{a+b}{c}+1+\frac{ab}{c^2}}\ge\frac{2ab}{c}+\left(a+b\right)\)\(\left(3\right)\).

Chứng minh tương tự, ta được:
\(\left(b+c\right)\sqrt{\frac{b+c}{a}+1+\frac{bc}{a^2}}\ge\frac{2bc}{a}+\left(b+c\right)\)\(\left(4\right)\).

Chứng minh tương tự, ta được:

\(\left(c+a\right)\sqrt{\frac{c+a}{b}+1+\frac{ca}{b^2}}\ge\frac{2ca}{b}+\left(c+a\right)\)\(\left(5\right)\).

Từ\(\left(3\right)\) , \(\left(4\right)\), \(\left(5\right)\), ta được:

\(\left(a+b\right)\sqrt{\frac{a+b}{c}+1+\frac{ab}{c^2}}+\left(b+c\right)\sqrt{\frac{b+c}{a}+1+\frac{bc}{a^2}}\)\(+\left(c+a\right)\sqrt{\frac{c+a}{b}+1+\frac{ca}{b^2}}\ge\frac{2ab}{c}+\left(a+b\right)\)\(+\frac{2bc}{a}+\left(b+c\right)+\frac{2ca}{b}+\left(c+a\right)\).

\(\Leftrightarrow A\ge\left(\frac{ab}{c}+\frac{bc}{a}\right)+\left(\frac{bc}{a}+\frac{ca}{b}\right)+\left(\frac{ca}{b}+\frac{ab}{c}\right)\)\(+2\left(a+b+c\right)\)\(\left(6\right)\).

Vì \(a,b,c>0\)nên áp dụng bất dẳng thức Cô-si cho 2 số dương, ta được:

\(\frac{ab}{c}+\frac{bc}{a}\ge2\sqrt{\frac{ab}{c}.\frac{bc}{a}}=2b\)\(\left(7\right)\).

Chứng minh tương tự, ta được:

\(\frac{bc}{a}+\frac{ca}{b}\ge2c\)\(\left(8\right)\).

Chứng minh tương tự, ta được:

\(\frac{ca}{b}+\frac{ab}{c}\ge2a\)\(\left(9\right)\).

Từ \(\left(7\right)\), \(\left(8\right)\), \(\left(9\right)\), ta được:

\(\left(\frac{ab}{c}+\frac{bc}{a}\right)+\left(\frac{bc}{a}+\frac{ca}{b}\right)+\left(\frac{ca}{b}+\frac{ab}{c}\right)\ge2a\)\(+2b+2c=2\left(a+b+c\right)\).

\(\Leftrightarrow\left(\frac{ab}{c}+\frac{bc}{a}\right)+\left(\frac{bc}{a}+\frac{ca}{b}\right)+\left(\frac{ca}{b}+\frac{ab}{c}\right)+\)\(2\left(a+b+c\right)\ge2\left(a+b+c\right)+2\left(a+b+c\right)\)\(=4\left(a+b+c\right)\)\(\left(10\right)\).

Từ \(\left(6\right)\)và \(\left(10\right)\), ta được:
\(A\ge4\left(a+b+c\right)\).

\(\Leftrightarrow A\ge4\sqrt{2021}\)(vì \(a+b+c=\sqrt{2021}\).

Dấu bằng xảy ra \(\Leftrightarrow\hept{\begin{cases}a=b=c>0\\a+b+c=\sqrt{2021}\end{cases}}\Leftrightarrow a=b=c=\frac{\sqrt{2021}}{3}\).

Vậy \(\sqrt{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\left(\frac{\sqrt{a+b}}{c}+\frac{\sqrt{b+c}}{a}+\frac{\sqrt{c+a}}{b}\right)\)đạt GTNN là \(4\sqrt{2021}\)\(\Leftrightarrow a=b=c=\frac{\sqrt{2021}}{3}\).

A B C 5 5 6 M N

a, Vì CN là phân giác ^C nên : \(\frac{AC}{BC}=\frac{AN}{NB}\)( t/c ) \(\Rightarrow\frac{AC}{AN}=\frac{BC}{NB}\)( tỉ lệ thức )

Vì BM là phân giác ^B nên : \(\frac{AB}{BC}=\frac{AM}{MC}\)( t/c ) \(\Rightarrow\frac{AB}{AM}=\frac{BC}{MC}\)( tỉ lệ thức )

mà \(AB=AC\)( do tam giác ABC cân ) suy ra : \(\frac{AB}{AM}=\frac{AC}{AN}\)

Vậy MN // BC ( theo talét đảo ) 

bổ sung hộ mình phần a là NB = MC ( do là phân giác mà tam giác ABC cân )

b, Xét tam giác ANC và tam giác AMB ta có : 

^A _ chung 

\(\frac{AC}{AN}=\frac{AB}{AM}\)( cma ) 

Vậy tam giác ANC ~ tam giác AMB ( c.g.c ) 

tim x hay thu gon

\(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\)

\(\Leftrightarrow\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}\)

\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)

\(\Leftrightarrow\frac{\left(x+7\right)}{\left(x+4\right)\left(x+7\right)}-\frac{\left(x+4\right)}{\left(x+4\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Leftrightarrow\frac{x+7-x-4}{\left(x+4\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Leftrightarrow\frac{3}{\left(x+4\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Leftrightarrow\left(x+4\right)\left(x+7\right)=54\)

\(\Leftrightarrow x^2+11x+28=54\)

\(\Leftrightarrow x^2+11x-26=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+13\right)=0\)

....tự lm tiếp

\(7x^2+12y^2=2013\)

\(12y^2⋮3,2013⋮3\Rightarrow7x^2⋮3\Rightarrow x^2⋮3\Rightarrow x⋮3\Rightarrow x=3a\).

\(63a^2+12y^2=2013\Leftrightarrow21x^2+4y^2=671\)

Ta có: \(y^2\)khi chia cho \(3\)chỉ dư \(0\)hoặc \(1\)nên \(4y^2\)chia cho \(3\)chỉ dư \(0\)hoặc \(1\).

mà \(671\equiv2\left(mod3\right),21x^2⋮3\) (mâu thuẫn) 

Do đó ta có đpcm.