\(a+bc=a\left(a+b+c\right)+bc=\left(a+b\right)\left(a+c\right)\)

\(\Rightarrow P=\dfrac{\left(a+b\right)\left(a+c\right)}{b+c}+\dfrac{\left(a+b\right)\left(b+c\right)}{a+c}+\dfrac{\left(a+c\right)\left(b+c\right)}{a+b}\)

Áp dụng BĐT Cô-si:

\(\dfrac{\left(a+b\right)\left(a+c\right)}{b+c}+\dfrac{\left(a+b\right)\left(b+c\right)}{a+c}\ge2\sqrt{\dfrac{\left(a+b\right)^2\left(a+c\right)\left(b+c\right)}{\left(b+c\right)\left(a+c\right)}}=2\left(a+b\right)\)

\(\dfrac{\left(a+b\right)\left(a+c\right)}{b+c}+\dfrac{\left(a+c\right)\left(b+c\right)}{a+b}\ge2\left(a+c\right)\)

\(\dfrac{\left(a+b\right)\left(b+c\right)}{a+c}+\dfrac{\left(a+c\right)\left(b+c\right)}{a+b}\ge2\left(a+b\right)\)

Cộng vế:

\(2P\ge4\left(a+b+c\right)=4\Rightarrow P\ge2\)

Cảm ơn thầy ạ.

a: \(\dfrac{x+4}{3}+\dfrac{x+3}{4}=\dfrac{x+2}{5}+\dfrac{x+1}{6}\)

=>\(\dfrac{x+4}{3}+1+\dfrac{x+3}{4}+1=\dfrac{x+2}{5}+1+\dfrac{x+1}{6}=1\)

=>\(\dfrac{x+7}{3}+\dfrac{x+7}{4}=\dfrac{x+7}{5}+\dfrac{x+7}{6}\)

=>\(\left(x+7\right)\left(\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{6}\right)=0\)

=>x+7=0

=>x=-7

b: \(\dfrac{x+1}{87}+\dfrac{x+3}{85}=\dfrac{x+5}{83}+\dfrac{x+7}{81}\)

=>\(\left(\dfrac{x+1}{87}+1\right)+\left(\dfrac{x+3}{85}+1\right)=\left(\dfrac{x+5}{83}+1\right)+\left(\dfrac{x+7}{81}+1\right)\)

=>\(\dfrac{x+88}{87}+\dfrac{x+88}{85}=\dfrac{x+88}{83}+\dfrac{x+88}{81}\)

=>\(\left(x+88\right)\left(\dfrac{1}{87}+\dfrac{1}{85}-\dfrac{1}{83}-\dfrac{1}{81}\right)=0\)

=>x+88=0

=>x=-88

c: \(\dfrac{x-29}{31}-\dfrac{x-27}{33}=\dfrac{x-17}{43}-\dfrac{x-15}{45}\)

=>\(\dfrac{x-29}{31}+\dfrac{x-15}{45}=\dfrac{x-17}{43}+\dfrac{x-27}{33}\)

=>\(\left(\dfrac{x-29}{31}-1\right)+\left(\dfrac{x-15}{45}-1\right)=\left(\dfrac{x-17}{43}-1\right)+\left(\dfrac{x-27}{33}-1\right)\)

=>\(\dfrac{x-60}{31}+\dfrac{x-60}{45}=\dfrac{x-60}{43}+\dfrac{x-60}{33}\)

=>\(\left(x-60\right)\left(\dfrac{1}{31}+\dfrac{1}{45}-\dfrac{1}{43}-\dfrac{1}{33}\right)=0\)

=>x-60=0

=>x=60

d: \(\dfrac{x+1}{11}-\dfrac{2x-5}{15}=\dfrac{3x-47}{17}-\dfrac{4x-59}{19}\)

=>\(\dfrac{x+1}{11}+\dfrac{4x-59}{19}=\dfrac{3x-47}{17}+\dfrac{2x-5}{15}\)

=>\(\dfrac{x+1}{11}-1+\dfrac{4x-59}{19}+1=\dfrac{3x-47}{17}+1+\dfrac{2x-5}{15}-1\)

=>\(\dfrac{x-10}{11}+\dfrac{4x-40}{19}=\dfrac{3x-30}{17}+\dfrac{2x-20}{15}\)

=>\(\left(x-10\right)\left(\dfrac{1}{11}+\dfrac{4}{19}-\dfrac{3}{17}-\dfrac{2}{15}\right)=0\)

=>x-10=0

=>x=10

ĐKXĐ: \(x\in\mathbb{R}\)

Đặt \(A=\dfrac{x+3}{x^2+7}\). Khi đó:

Xét: \(A-\dfrac{1}{2}=\dfrac{x+3}{x^2+7}-\dfrac{1}{2}=\dfrac{2\left(x+3\right)}{2\left(x^2+7\right)}-\dfrac{x^2+7}{2\left(x^2+7\right)}\)

\(=\dfrac{2x+6-x^2-7}{2\left(x^2+7\right)}=\dfrac{-x^2+2x-1}{2\left(x^2+7\right)}=\dfrac{-\left(x-1\right)^2}{2\left(x^2+7\right)}\)

Ta thấy: \(\left\{{}\begin{matrix}\left(x-1\right)^2\ge0\forall x\\2\left(x^2+7\right)>0\forall x\end{matrix}\right.\Rightarrow\dfrac{\left(x-1\right)^2}{2\left(x^2+7\right)}\ge0\)

\(\Rightarrow\dfrac{-\left(x-1\right)^2}{2\left(x^2+7\right)}\le0\)

\(\Leftrightarrow A-\dfrac{1}{2}\le0\Leftrightarrow A\le\dfrac{1}{2}\)

Dấu \("="\) xảy ra khi: \(x-1=0\Leftrightarrow x=1\)

Vậy GTLN của biểu thức đã cho là \(\dfrac{1}{2}\) tại \(x=1\).

\(\text{#}Toru\)

wow

a: ΔABC vuông tại A

=>\(S_{ABC}=\dfrac{1}{2}\cdot AB\cdot AC\left(1\right)\)

Ta có: ΔABC có AH là đường cao

=>\(S_{ABC}=\dfrac{1}{2}\cdot AH\cdot BC\left(2\right)\)

Từ (1) và (2) suy ra \(AB\cdot AC=AH\cdot BC\)

b: Xét ΔAMH vuông tại M và ΔAHB vuông tại H có

\(\widehat{MAH}\) chung

Do đó: ΔAMH~ΔAHB

=>\(\dfrac{AM}{AH}=\dfrac{AH}{AB}\)

=>\(AH^2=AM\cdot AB\left(3\right)\)

Xét ΔANH vuông tại N và ΔAHC vuông tại H có

\(\widehat{NAH}\) chung

Do đó: ΔANH~ΔAHC

=>\(\dfrac{AN}{AH}=\dfrac{AH}{AC}\)

=>\(AH^2=AN\cdot AC\left(4\right)\)

Từ (3) và (4) suy ra \(AM\cdot AB=AN\cdot AC\)

=>\(\dfrac{AM}{AC}=\dfrac{AN}{AB}\)

Xét ΔAMN và ΔACB có

\(\dfrac{AM}{AC}=\dfrac{AN}{AB}\)

\(\widehat{MAN}\) chung

Do đó: ΔAMN~ΔACB

a: Ta có: \(BM=MA=\dfrac{BA}{2}\)

\(BN=NC=\dfrac{BC}{2}\)

mà BA=BC

nên BM=MA=BN=NC

Xét ΔMBC vuông tại B và ΔNCD vuông tại C có

BC=CD

MB=NC

Do đó: ΔMBC=ΔNCD

=>\(\widehat{BMC}=\widehat{CND}\)

mà \(\widehat{BMC}+\widehat{NCI}=90^0\)(ΔBCM vuông tại B)

nên \(\widehat{CND}+\widehat{NCI}=90^0\)

=>CM\(\perp\)DN tại I

=>ΔEID vuông tại I

anh ơi còn câu b giúp em nốt vs ạ

Bài 3:

Xét ΔABC có \(CA^2+CB^2=AB^2\)

nên ΔABC vuông tại C

=>\(S_{ACB}=\dfrac{1}{2}\cdot CA\cdot CB=\dfrac{1}{2}\cdot12\cdot16=6\cdot16=96\left(cm^2\right)\)

Bài 5:

Xét ΔMNP có \(MN^2+MP^2=NP^2\)

nên ΔMNP vuông tại M

Xét ΔMNP vuông tại M có MK là đường cao

nên \(MK\cdot NP=MN\cdot MP\)

=>\(MK\cdot25=15\cdot20=300\)

=>\(MK=\dfrac{300}{25}=12\left(cm\right)\)

a: \(\dfrac{2x}{5}+\dfrac{2}{5}=\dfrac{2x+2}{5}\)

b: \(\dfrac{2x+5}{3}+\dfrac{x-2}{3}\)

\(=\dfrac{2x+5+x-2}{3}\)

\(=\dfrac{3x+3}{3}=x+1\)

c: ĐKXĐ: \(x\ne1\)

\(\dfrac{5x}{x-1}+\dfrac{x+1}{x-1}\)

\(=\dfrac{5x+x+1}{x-1}\)

\(=\dfrac{6x+1}{x-1}\)

Xét ΔBEH vuông tại E và ΔBDA vuông tại D có

\(\widehat{EBH}\) chung

Do đó: ΔBEH~ΔBDA

=>\(\dfrac{BE}{BD}=\dfrac{BH}{BA}\)

=>\(BE\cdot BA=BH\cdot BD\)