4)Giải các phương trình
a) 3x+22−3x+16=53+2x b)(x+1)(x+2)=(2-x)(x+2)
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3 tháng 2
c: \(3a^2-6ab+3b^2-12c^2\)
\(=3\left(a^2-2ab+b^2-4c^2\right)\)
\(=3\left[\left(a-b\right)^2-\left(2c\right)^2\right]\)
\(=3\left(a-b-2c\right)\left(a-b+2c\right)\)
d: \(x^2-25+y^2+2xy\)
\(=\left(x^2+2xy+y^2\right)-25\)
\(=\left(x+y\right)^2-25\)
=(x+y+5)(x+y-5)
3 tháng 2
\(A=\dfrac{x^4-\left(x-1\right)^2}{\left(x^2+1\right)^2-x^2}+\dfrac{x^2-\left(x^2-1\right)^2}{x^2\left(x+1\right)^2-1}+\dfrac{x^2\left(x-1\right)^2-1}{x^4-\left(x+1\right)^2}\)
\(=\dfrac{\left(x^2-x+1\right)\left(x^2+x-1\right)}{\left(x^2+1-x\right)\left(x^2+1+x\right)}+\dfrac{\left(x-x^2+1\right)\left(x+x^2-1\right)}{\left(x^2+x-1\right)\left(x^2+x+1\right)}+\dfrac{\left(x^2-x-1\right)\left(x^2-x+1\right)}{\left(x^2-x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x^2+x-1}{x^2+x+1}+\dfrac{-x^2+x+1}{x^2+x+1}+\dfrac{x^2-x+1}{x^2+x+1}\)
\(=\dfrac{x^2+x-1-x^2+x+1+x^2-x+1}{x^2+x+1}\)
\(=\dfrac{x^2+x+1}{x^2+x+1}=1\)
3 tháng 2
\(B=\dfrac{1}{a-b}+\dfrac{1}{a+b}+\dfrac{2a}{a^2+b^2}+\dfrac{4a^3}{a^4+b^4}+\dfrac{8a^7}{a^8+b^8}\)
\(=\dfrac{a+b+a-b}{\left(a-b\right)\left(a+b\right)}+\dfrac{2a}{a^2+b^2}+\dfrac{4a^3}{a^4+b^4}+\dfrac{8a^7}{a^8+b^8}\)
\(=\dfrac{2a}{a^2-b^2}+\dfrac{2a}{a^2+b^2}+\dfrac{4a^3}{a^4+b^4}+\dfrac{8a^7}{a^8+b^8}\)
\(=\dfrac{2a\left(a^2+b^2\right)+2a\left(a^2-b^2\right)}{\left(a^2-b^2\right)\left(a^2+b^2\right)}+\dfrac{4a^3}{a^4+b^4}+\dfrac{8a^7}{a^8+b^8}\)
\(=\dfrac{4a^3}{a^4-b^4}+\dfrac{4a^3}{a^4+b^4}+\dfrac{8a^7}{a^8+b^8}\)
\(=\dfrac{4a^3\left(a^4+b^4+a^4-b^4\right)}{\left(a^4-b^4\right)\left(a^4+b^4\right)}+\dfrac{8a^7}{a^8+b^8}\)
\(=\dfrac{8a^7}{a^8-b^8}+\dfrac{8a^7}{a^8+b^8}\)
\(=\dfrac{8a^7\left(a^8-b^8+a^8+b^8\right)}{a^{16}-b^{16}}=\dfrac{16a^{15}}{a^{16}-b^{16}}\)
2 tháng 6
Ta có: \(\left(a+b+c\right)^2=a^2+b^2+c^2\)
=>\(a^2+b^2+c^2+2\left(ab+ac+bc\right)=a^2+b^2+c^2\)
=>2(ab+ac+bc)=0
=>2ab+2ac+2bc=0
=>ab+ac+bc=0
=>ab=-ac-bc; ac=-ab-bc; bc=-ab-ac
\(a^2+2bc=a^2+bc+bc=a^2+bc-ab-ac\)
\(=a^2-ab-ac+bc=a\left(a-b\right)-c\left(a-b\right)=\left(a-b\right)\left(a-c\right)\)
\(b^2+2ac=b^2+ac+ac=b^2+ac-ba-bc\)
\(=b^2-ba-bc+ac=b\left(b-a\right)-c\left(b-a\right)\)
=(b-c)(b-a)=-(a-b)(b-c)
\(c^2+2ab=c^2+ab+ab=c^2+ab-ac-bc\)
\(=c^2-ac-bc+ab=c\left(c-a\right)-b\left(c-a\right)=\left(c-a\right)\left(c-b\right)=\left(a-c\right)\left(b-c_{}\right)\)
\(P=\frac{2}{a^2+2bc}+\frac{2}{b^2+2ac}+\frac{2}{c^2+2ab}\)
\(=\frac{2}{\left(a-b\right)\left(a-c\right)}-\frac{2}{\left(a-b\right)\left(b-c\right)}+\frac{2}{\left(a-c\right)\cdot\left(b-c\right)}\)
\(=\frac{2\left(b-c\right)-2\left(a-c\right)+2\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=\frac{2\left(b-c-a+c+a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=0\)
3 tháng 2
Cứu mình với !!!😭Năm nay mình học lớp 6,nhưng vượt cấp lớp 8 IOE nên có không biết câu sắp xếp này,bạn nào biết thì giúp mình với !Mình xin cảm ơn
a: 3x+22-3x+16=53+2x
=>2x+53=38
=>2x=38-53=-15
=>\(x=-\dfrac{15}{2}\)
b: \(\left(x+1\right)\left(x+2\right)=\left(2-x\right)\left(x+2\right)\)
=>\(\left(x+1\right)\left(x+2\right)-\left(2-x\right)\left(x+2\right)=0\)
=>\(\left(x+2\right)\left(x+1-2+x\right)=0\)
=>(x+2)(2x-1)=0
=>\(\left[{}\begin{matrix}x+2=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{2}\end{matrix}\right.\)