That's.....question to answer (easy)
This textbooks is....to undersstand
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________, I decided to stop trading with them
A.Despite of the fact that they were the biggest dealer
B.Though being the biggest dealer
C.Being the biggest dealer
D.Even though they were the biggest dealer
a,b,c là các số thực đôi một phân biệt
=>\(a-b;b-c;a-c\) đều khác 0
\(a^3+b^3+c^3=3bac\)
=>\(\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
=>\(\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)=0\)
=>\(\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\)
=>\(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)
=>\(\left(a+b+c\right)\left[2a^2+2b^2+2c^2-2ab-2ac-2bc\right]=0\)
=>\(\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\right]=0\)
=>\(\left[{}\begin{matrix}a+b+c=0\\\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a=b=c\left(loại\right)\end{matrix}\right.\)
=>a+b+c=0
=>a+b=-c; a+c=-b; b+c=-a
\(P=\dfrac{a+b}{c}\cdot\dfrac{b+c}{a}\cdot\dfrac{c+a}{b}=\dfrac{-c}{c}\cdot\dfrac{-a}{a}\cdot\dfrac{-b}{b}=-1\)
a: ĐKXĐ: \(x\notin\left\{1;-1\right\}\)
\(C=\left(\dfrac{1}{x^2+1}-\dfrac{x+1}{x^4-1}\right):\dfrac{x+1}{x^5+x^4-x-1}\)
\(=\dfrac{x^2-1-x-1}{\left(x^2+1\right)\left(x^2-1\right)}:\dfrac{x+1}{x^4\left(x+1\right)-\left(x+1\right)}\)
\(=\dfrac{x^2-x-2}{\left(x^2+1\right)\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x+1\right)\left(x^4-1\right)}{x+1}\)
\(=\dfrac{\left(x-2\right)\left(x+1\right)}{\left(x^2+1\right)\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x^4-1}{1}\)
=(x-2)(x+1)
b: Để C=0 thì (x-2)(x+1)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(nhận\right)\\x=-1\left(loại\right)\end{matrix}\right.\)
c: \(C=\left(x-2\right)\left(x+1\right)=x^2-x-2\)
\(=x^2-x+\dfrac{1}{4}-\dfrac{9}{4}\)
\(=\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{4}>=-\dfrac{9}{4}\forall x\)
Dấu '=' xảy ra khi \(x-\dfrac{1}{2}=0\)
=>\(x=\dfrac{1}{2}\)
a: Xét tứ giác ABCD có \(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^0\)
=>\(\widehat{C}+\widehat{D}=360^0-110^0-70^0=180^0\)
=>\(\dfrac{1}{3}\cdot\widehat{D}+\widehat{D}=180^0\)
=>\(\dfrac{4}{3}\cdot\widehat{D}=180^0\)
=>\(\widehat{D}=135^0\)
\(\widehat{C}=\dfrac{1}{3}\cdot135^0=45^0\)
b:
Sửa đề: Cho tứ giác ABCD.
Đặt \(\widehat{B}=x;\widehat{C}=y;\widehat{D}=z\)
\(\dfrac{\widehat{B}}{2}=\dfrac{\widehat{C}}{3}=\dfrac{\widehat{D}}{4}\)
=>\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)
Xét tứ giác ABCD có \(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^0\)
=>\(x+y+z=360^0-90^0=270^0\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+z}{2+3+4}=\dfrac{270}{9}=30^0\)
=>\(x=2\cdot30^0=60^0;y=3\cdot30^0=90^0;z=4\cdot30^0=120^0\)
Vậy: \(\widehat{B}=x=60^0;\widehat{C}=y=90^0;\widehat{D}=z=120^0\)
\(\widehat{C}=\widehat{B}+10^0=\widehat{A}+10^0+10^0=\widehat{A}+20^0\)
\(\widehat{D}=\widehat{C}+10^0=\widehat{A}+20^0+10^0=\widehat{A}+30^0\)
Xét tứ giác ABCD có \(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^0\)
=>\(\widehat{A}+\widehat{A}+10^0+\widehat{A}+20^0+\widehat{A}+30^0=360^0\)
=>\(4\cdot\widehat{A}=300^0\)
=>\(\widehat{A}=75^0\)
\(\widehat{B}=75^0+10^0=85^0\)
\(\widehat{C}=75^0+20^0=95^0\)
\(\widehat{D}=75^0+30^0=105^0\)
\(\left(x+y\right)^2-2\left(x+y\right)+1\)
\(=\left(x+y-1\right)^2\) (HĐT số 2)
bài này ngữ pháp là gì
That's the easiest question to answer
This textbook is easy to understand