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\(\dfrac{-2}{3}+\left(\dfrac{3}{4}-x\right)=\dfrac{4}{5}\\ \Rightarrow\dfrac{3}{4}-x=\dfrac{4}{5}+\dfrac{2}{3}=\dfrac{22}{15}\\ \Rightarrow x=\dfrac{3}{4}-\dfrac{22}{15}=-\dfrac{43}{60}\)
ΔAED vuông tại E nên AE<AD
ΔDFC vuông tại F nên FC<DC
=>AE+FC<AD+DC=AC
Bài 2:
a: \(f\left(x\right)=-9x^3-2x^2+6x-3\)
\(G\left(x\right)=9x^3-6x+53\)
b: \(H\left(x\right)=9x^3-6x+53-9x^3-2x^2+6x-3=-2x^2+50\)
c: Đặt H(x)=0
=>2x2-50=0
=>x=5 hoặc x=-5
\(\dfrac{3}{4}-\left(x-\dfrac{1}{2}\right)=\dfrac{5}{3}\)
\(\Leftrightarrow\dfrac{3}{4}-x+\dfrac{1}{2}=\dfrac{5}{3}\)
\(\Leftrightarrow x=\dfrac{3}{4}+\dfrac{1}{2}-\dfrac{5}{3}\)
\(\Leftrightarrow x=-\dfrac{5}{12}\)
Ta có :
\(\frac{x-3}{97}+\frac{x-27}{73}+\frac{x-67}{33}+\frac{x-73}{27}=4\)
\(\Leftrightarrow\left(\frac{x-3}{97}-1\right)+\left(\frac{x-27}{73}-1\right)+\left(\frac{x-67}{33}-1\right)+\left(\frac{x-73}{27}-1\right)=0\)
\(\Leftrightarrow\frac{x-100}{97}+\frac{x-100}{73}+\frac{x-100}{33}+\frac{x-100}{27}=0\)
\(\Leftrightarrow\left(x-100\right)\left(\frac{1}{97}+\frac{1}{73}+\frac{1}{33}+\frac{1}{27}\right)=0\)
Vì \(\frac{1}{97}+\frac{1}{73}+\frac{1}{33}+\frac{1}{27}>0\) Nên \(x-100=0\)
\(\Leftrightarrow x=100\)
Vậy \(x=100\)
\(\Leftrightarrow\frac{x-3}{87}+\frac{x-27}{79}+\frac{x-67}{33}+\frac{x-73}{27}-4=0\)
\(\Leftrightarrow\left(\frac{x-3}{97}-1\right)+\left(\frac{x-27}{73}-1\right)+\left(\frac{x-67}{33}-1\right)+\left(\frac{x-73}{27}-1\right)=0\)
\(\Leftrightarrow\left(\frac{x-3-97}{97}\right)+\left(\frac{x-27-73}{73}\right)+\left(\frac{x-67-33}{33}\right)+\left(\frac{x-73-27}{27}\right)=0\)
\(\Leftrightarrow\frac{x-100}{97}+\frac{x-100}{73}+\frac{x-100}{33}+\frac{x-100}{27}=0\)
\(\Leftrightarrow\left(x-100\right)\left(\frac{1}{97}+\frac{1}{73}+\frac{1}{33}+\frac{1}{27}\right)=0\)
Vì \(\frac{1}{97}+\frac{1}{73}+\frac{1}{33}+\frac{1}{27}\ne0\)
\(\Rightarrow x-100=0\Leftrightarrow x=100\)
Chu vi=20cm=>AB+AC=20-6=14=>Cân=>AB=AC=14/2=7
BC<AB=AC
=>A<C=B
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