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|x|+|y|<=3
mà x,y nguyên
nên \(\left(\left|x\right|;\left|y\right|\right)\in\left\{\left(0;3\right);\left(0;1\right);\left(0;2\right);\left(0;0\right);\left(1;1\right);\left(1;2\right);\left(3;0\right);\left(1;0\right);\left(2;0\right);\left(2;1\right)\right\}\)
=>(x;y)\(\in\){(0;0);(0;1);(1;0);(0;-1);(-1;0);(0;2);(2;0);(0;-2);(-2;0);(0;3);(0;-3);(3;0);(-3;0);(1;1);(1;-1);(-1;1);(1;2);(2;1);(-1;-2);(-2;-1);(1;-2);(-2;1);(-1;2);(2;-1)}
Xét ΔABC có \(\widehat{ABC}+\widehat{ACB}+\widehat{BAC}=180^0\)
=>\(2\cdot\left(\widehat{IBC}+\widehat{ICB}\right)=180^0-\widehat{BAC}\)
=>\(\widehat{IBC}+\widehat{ICB}=90^0-\dfrac{1}{2}\cdot\widehat{BAC}\)
Xét ΔBIC có \(\widehat{BIC}+\widehat{IBC}+\widehat{ICB}=180^0\)
=>\(\widehat{BIC}+90^0-\dfrac{1}{2}\widehat{BAC}=180^0\)
=>\(\widehat{BIC}=180^0-90^0+\dfrac{1}{2}\cdot\widehat{BAC}=90^0+\dfrac{1}{2}\cdot\widehat{BAC}\)
\(-\dfrac{2}{5}+\dfrac{3}{4}-\dfrac{-1}{6}+\dfrac{-2}{5}\\ =\left(\dfrac{-2}{5}+\dfrac{-2}{5}\right)+\left(\dfrac{3}{4}+\dfrac{1}{6}\right)\\ =\dfrac{-4}{5}+\left(\dfrac{9}{12}+\dfrac{2}{12}\right)\\ =\dfrac{-4}{5}+\dfrac{11}{12}\\ =\dfrac{-48}{60}+\dfrac{55}{60}\\ =\dfrac{55-48}{60}\\ =\dfrac{7}{60}\)
Bài 10:
Đặt \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=k\)
=>\(\left\{{}\begin{matrix}c=dk\\b=ck=dk\cdot k=dk^2\\a=bk=dk^2\cdot k=dk^3\end{matrix}\right.\)
a:
\(\left(\dfrac{a+b+c}{b+c+d}\right)^3=\left(\dfrac{dk^3+dk^2+dk}{dk^2+dk+d}\right)^3=\left(\dfrac{dk\left(k^2+k+1\right)}{d\left(k^2+k+1\right)}\right)^3=k^3\)
\(\dfrac{a}{d}=\dfrac{dk^3}{d}=k^3\)
Do đó: \(\dfrac{a}{d}=\left(\dfrac{a+b+c}{b+c+d}\right)^3\)
b: Sửa đề: Chứng minh \(\dfrac{a^3+c^3+b^3}{c^3+b^3+d^3}=\dfrac{a}{d}\)
\(\dfrac{a^3+c^3+b^3}{c^3+b^3+d^3}=\dfrac{\left(dk^3\right)^3+\left(dk\right)^3+\left(dk^2\right)^3}{\left(dk\right)^3+\left(dk^2\right)^3+d^3}\)
\(=\dfrac{d^3k^9+d^3k^3+d^3k^6}{d^3k^3+d^3k^6+d^3}=\dfrac{d^3\cdot k^3\left(k^6+1+k^3\right)}{d^3\cdot\left(k^3+k^6+1\right)}=k^3\)
\(=\dfrac{dk^3}{d}=\dfrac{a}{d}\)
Bài 14:
x+y+z=0
=>x+y=-z; x+z=-y; y+z=-x
\(A=\left(x+y\right)\left(y+z\right)\left(x+z\right)=\left(-z\right)\cdot\left(-x\right)\cdot\left(-y\right)\)
=-xyz
=-2
Bài 10:
Đặt \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=k\)
=>\(\left\{{}\begin{matrix}c=dk\\b=ck=dk\cdot k=dk^2\\a=bk=dk^2\cdot k=dk^3\end{matrix}\right.\)
a:
\(\left(\dfrac{a+b+c}{b+c+d}\right)^3=\left(\dfrac{dk^3+dk^2+dk}{dk^2+dk+d}\right)^3=\left(\dfrac{dk\left(k^2+k+1\right)}{d\left(k^2+k+1\right)}\right)^3=k^3\)
\(\dfrac{a}{d}=\dfrac{dk^3}{d}=k^3\)
Do đó: \(\dfrac{a}{d}=\left(\dfrac{a+b+c}{b+c+d}\right)^3\)
b: Sửa đề: Chứng minh \(\dfrac{a^3+c^3+b^3}{c^3+b^3+d^3}=\dfrac{a}{d}\)
\(\dfrac{a^3+c^3+b^3}{c^3+b^3+d^3}=\dfrac{\left(dk^3\right)^3+\left(dk\right)^3+\left(dk^2\right)^3}{\left(dk\right)^3+\left(dk^2\right)^3+d^3}\)
\(=\dfrac{d^3k^9+d^3k^3+d^3k^6}{d^3k^3+d^3k^6+d^3}=\dfrac{d^3\cdot k^3\left(k^6+1+k^3\right)}{d^3\cdot\left(k^3+k^6+1\right)}=k^3\)
\(=\dfrac{dk^3}{d}=\dfrac{a}{d}\)
Bài 8:
a: \(\dfrac{x}{5}=\dfrac{y}{6}\)
=>\(\dfrac{x}{20}=\dfrac{y}{24}\)
\(\dfrac{y}{8}=\dfrac{z}{7}\)
=>\(\dfrac{y}{24}=\dfrac{z}{21}\)
Do đó: \(\dfrac{x}{20}=\dfrac{y}{24}=\dfrac{z}{21}=k\)
=>x=20k;y=24k;z=21k
x+y-z=69
=>20k+24k-21k=69
=>23k=69
=>k=3
=>\(x=20\cdot3=60;y=24\cdot3=72;z=21\cdot3=63\)
b: Đặt \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=k\)
=>x=3k;y=4k;z=5k
\(2x^2+2y^2-3z^2=-100\)
=>\(2\cdot\left(3k\right)^2+2\cdot\left(4k\right)^2-3\cdot\left(5k\right)^2=-100\)
=>\(k^2=4\)
=>\(\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\)
TH1: k=2
=>\(x=3\cdot2=6;y=4\cdot2=8;z=5\cdot2=10\)
TH2: k=-2
=>\(x=3\cdot\left(-2\right)=-6;y=4\cdot\left(-2\right)=-8;z=5\cdot\left(-2\right)=-10\)
a, Theo tc dãy tỉ số bằng nhau
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{x-y}{3-5}=\dfrac{22}{-2}=-11\Rightarrow x=-33;y=-55\)
b, \(\dfrac{5}{2}=\dfrac{y}{x}\Rightarrow\dfrac{x}{2}=\dfrac{y}{5}\)Theo tc dãy tỉ số bằng nhau
\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=-\dfrac{21}{7}=-3\Rightarrow x=-6;y=-15\)
c, \(7x=4y\Rightarrow\dfrac{x}{4}=\dfrac{y}{7}\)Theo tc dãy tỉ số bằng nhau
\(\dfrac{x}{4}=\dfrac{y}{7}=\dfrac{x-y}{3-7}=-\dfrac{21}{-4}=7\Rightarrow x=28;y=49\)
Bài 11: \(\dfrac{2a+13b}{3a-7b}=\dfrac{2c+13d}{3c-7d}\)
=>\(\left(2a+13b\right)\left(3c-7d\right)=\left(3a-7b\right)\left(2c+13d\right)\)
=>\(6ac-14ad+39bc-91bd=6ac+39ad-14bc-91bd\)
=>-14ad-39ad=-14bc-39bc
=>ad=bc
=>\(\dfrac{a}{b}=\dfrac{c}{d}\)
Bài 12:
\(\dfrac{a+2019}{a-2019}=\dfrac{b+2020}{b-2020}\)
=>\(\left(a+2019\right)\left(b-2020\right)=\left(a-2019\right)\left(b+2020\right)\)
=>\(ab-2020a+2019b-2019\cdot2020=ab+2020a-2019b-2019\cdot2020\)
=>-2020a-2020a=-2019b-2019b
=>2020a=2019b
=>\(\dfrac{a}{2019}=\dfrac{b}{2020}\)
Bài 1:
a: \(\dfrac{a}{b}>1\)
=>\(\dfrac{a}{b}-1>0\)
=>\(\dfrac{a-b}{b}>0\)
mà b>0
nên a-b>0
=>a>b
b: a>b
=>\(\dfrac{a}{b}>\dfrac{b}{b}\)
=>\(\dfrac{a}{b}>1\)
c: a/b<1
=>\(\dfrac{a}{b}-1< 0\)
=>\(\dfrac{a-b}{b}< 0\)
mà b>0
nên a-b<0
=>a<b
d: a<b
=>\(\dfrac{a}{b}< \dfrac{b}{b}\)
=>\(\dfrac{a}{b}< 1\)