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a: M+N-P
\(=7a^2-2a+1-a^2+4\)
\(=6a^2-2a+5\)
b: \(=2y-x-2x+y+y+3x-5y+x\)
\(=-3x+3y-4y+4x=x-y\)
\(=a^2+2ab+b^2-a^2+2ab-b^2=4ab\)
c: \(=\left[{}\begin{matrix}5x-3-2x+1=3x-2\left(x>=\dfrac{1}{2}\right)\\5x-3+2x-1=7x-4\left(x< \dfrac{1}{2}\right)\end{matrix}\right.\)
mk làm câu a thôi, b dài nhưng tương tự
Gọi a/b=c/d=k =>a=bk ; c=dk
=>\(\frac{\left(2a+3b\right)^2}{\left(3a-4b\right)^2}=\frac{\left(2bk+3b\right)^2}{\left(3bk-4b\right)^2}=\frac{\left[b\left(2k+3\right)\right]^2}{\left[b\left(3k-4\right)\right]^2}=\frac{b^2\left(2k+3\right)^2}{b^2\left(3k-4\right)^2}=\frac{\left(2k+3\right)^2}{\left(3k-4\right)^2}\)(1)
=>\(\frac{\left(2c+3d\right)^2}{\left(3c-4d\right)^2}=\frac{\left(2dk+3d\right)^2}{\left(3dk-4d\right)^2}=\frac{\left[d\left(2k+3\right)\right]^2}{\left[d\left(3k-4\right)\right]^2}=\frac{\left(2k+3\right)^2}{\left(3k-4\right)^2}\)(2)
Từ (1);(2)=> đpcm
\(S=\frac{2a+2ab-b-1}{3b\left(2a-1\right)+6a-3}\\ =\frac{2a\left(b+1\right)-\left(b+1\right)}{3b\left(2a-1\right)+3\left(2a-1\right)}\\ =\frac{\left(2a-1\right)\left(b+1\right)}{3\left(b+1\right)\left(2a-1\right)}\\=\frac{1}{3}\)
Đặt \(\hept{\begin{cases}a-b=x\\b-c=y\\c-a=z\end{cases}}\)
\(A=\frac{2}{x}+\frac{2}{y}+\frac{2}{z}+\frac{x^2y^2z^2}{xyz}\)
\(A=\frac{\left(2y+2x\right).z+2xy}{xyz}+\frac{x^2+y^2+x^2}{xyz}\)
\(A=\frac{2yz+2xz+2xy}{xyz}+\frac{x^2+y^2+z^2}{xyz}\)
\(A=\frac{2yz+2xz+2xy+x^2+y^2+z^2}{xyz}=\frac{\left(x+y+z\right)^2}{xyz}\)
Có đúng k nhỉ k chắc
Đặt \(\hept{\begin{cases}a+b=m\\b+c=n\\c+a=p\end{cases}}\)
Xem VT = A
\(\Rightarrow A=m^2+n^2+p^2-mn-np-mp\)
\(2A=\left(m-n\right)^2+\left(n-p\right)^2+\left(p-m\right)^2\)
\(=\left(a+b-b-c\right)^2+\left(b+c-c-a\right)^2+\left(c+a-a-b\right)^2\)
\(=\left(a-c\right)^2+\left(b-a\right)^2+\left(c-b\right)^2\)
\(=a^2-2ac+c^2+b^2-2ab+a^2+c^2-2bc+b^2\)
\(=2\left(a^2+b^2+c^2-2ab-2bc-2ac\right)\)
\(\Rightarrow A=a^2+b^2+c^2-ab-bc-ca\)(đpcm)
Ta có:\(\frac{3a+b+c+d}{a}=\frac{a+3b+c+d}{b}=\frac{a+b+3c+d}{c}=\frac{a+b+c+3d}{d}\)
\(\Rightarrow\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}\)
\(\Rightarrow\orbr{\begin{cases}a+b+c+d=0\\a=b=c=d\end{cases}}\)
\(TH1:a+b+c+d=0\Rightarrow\hept{\begin{cases}a+b=-\left(c+d\right)\\b+c=-\left(a+d\right)\end{cases}}\)
\(\Rightarrow Q=\left(\frac{-\left(c+d\right)}{c+d}\right)^2+\left(\frac{-\left(a+d\right)}{a+d}\right)^2+\left(\frac{c+d}{-\left(c+d\right)}\right)^2+\left(\frac{a+d}{-\left(a+d\right)}\right)^2\)
\(\Rightarrow Q=\left(-1\right)^2\cdot4=1\cdot4=4\)
\(TH2:a=b=c=d\)
\(\Rightarrow Q=\left(\frac{a+a}{a+a}\right)^2+\left(\frac{a+a}{a+a}\right)^2+\left(\frac{a+a}{a+a}\right)^2+\left(\frac{a+a}{a+a}\right)^2=1^2\cdot4=1\cdot4=4\)
Vậy Q=4
Ta có:
\(2\left(a^2+b^2\right)=5ab\)
\(\Leftrightarrow2a^2-5ab+2b^2=0\)
\(\Leftrightarrow2a^2-4ab-ab+2b^2=0\)
\(\Leftrightarrow2a\left(a-2b\right)-b\left(a-2b\right)=0\)
\(\Leftrightarrow\left(2a-b\right)\left(a-2b\right)=0\)
\(\Leftrightarrow a=2b\) hay \(b=2a\)
Vì \(a>b>c\Leftrightarrow a=2b\)
\(\Leftrightarrow\frac{3a-b}{2a+b}=\frac{3.2b-b}{2.2b+b}=\frac{5b}{5b}=1\)
Vậy \(\frac{3a-b}{2a+b}=1\)