2m² -20m +14 Giúp mình phân tích thành hằng đẳng thức để tìm GTNN với
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\(x^6-y^6\)
\(=\left(x^3-y^3\right)\left(x^3+y^3\right)\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2-xy+y^2\right)\)
hk
tốt
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\(\sqrt[3]{15\sqrt{3}-26}=\sqrt[3]{-\left(26-15\sqrt{3}\right)}\)
\(=-\sqrt[3]{8-3\cdot2^2\cdot\sqrt{3}+3\cdot2\cdot3-3\sqrt{3}}\)
\(=-\sqrt[3]{\left(2-\sqrt{3}\right)^3}=-\left(2-\sqrt{3}\right)=-2+\sqrt{3}\)
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\(\left(a+b\right)^6=a^6+6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5+b^6
\)
Học dỏi nha
~ Good luck ~
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\(12-3\sqrt{12}=9-\sqrt{108}+3=9-2\sqrt{27}+3=\left(3-\sqrt{3}\right)^2\)
Bạn ei là hàng đẳng thức \(\left(a-b\right)^2\)??
\(12-3\sqrt{12}=12-3\sqrt{4.3}=12-3.2.\sqrt{3}\)
\(=9-2.3.\sqrt{3}+3=3^2-2.3.\sqrt{3}+\left(\sqrt{3}\right)^2\)
=\(\left(3-\sqrt{3}\right)^2\)
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\(\left(x-2\right)^3-1=\left(x-2\right)\left[\left(x-3\right)^2+x-2\right]=\left(x-2\right)\left(x^2+5x+7\right)\)
\(\left(x+3y\right)^2-9y^2=x\left(x+6y\right)\)
\(\left(x+3\right)^2-\left(x-1\right)^2=4\left(2x+4\right)=8\left(x+2\right)\)
a) \(\left(x-2\right)^3-1=\left(x-2\right)^3-1^3=\left(x-2-1\right)\left[\left(x-2\right)^2+\left(x-2\right)\cdot1+1^2\right]\)\(=\left(x-3\right)\left(x^2-4x+4+x-2+1\right)\)
\(=\left(x-3\right)\left(x^2-3x+3\right)\)
b) \(\left(x+3y\right)^2-9y^2\)
\(=\left(x+3y\right)^2-\left(3y\right)^2\)
\(=\left(x+3y+3y\right)\left(x+3y-3y\right)\)
\(=x\left(x+6y\right)\)
c) \(\left(x+3\right)^2-\left(x-1\right)^2\)
\(=\left(x+3-x+1\right)\left(x+3+x-1\right)\)
\(=4\left(2x+2\right)\)
\(=8\left(x+1\right)\)
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\(\left(c^2+b^2-5\right)^2-4\left(ab+2\right)^2\)
\(=\left(a^2+b^2-5-2ab-4\right)\left(a^2+b^2-5+2ab+4\right)\)
\(=\left(a^2+b^2-2ab-9\right)\left(a^2+b^2+2ab-1\right)\)
\(=\left[\left(a-b\right)^2-9\right]\left[\left(a+b\right)^2-1\right]\)
\(=\left(a-b-3\right)\left(a-b+3\right)\left(a+b-1\right)\left(a+b+1\right)\)
Phân tích thành nhân tử : -4x^2+12xy-9y^2+25(sử dụng hằng đẳng thức)
giúp mink với, mink đang cần gấp
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\(=-2x^2+6xy-3y^2+25\)
\(=-\left(2x^2-6xy+3y^2\right)+25\)
\(=-\left(2x^2+3y^2\right)+25\)
\(=\left(2x^2-3y^2\right)+25\)
\(=\left(2x^2+3y^2\right).\left(2x^2-3y^2\right)+25\)
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= (3x + 1 - x - 1)(3x + 1 + x + 1)
= 2x(4x + 2)
Em áp dụng hđt số 3 trong sgk nhé.
\(=2\left(m^2-10m+7\right)=2\left(m^2-10m+25-18\right)\)
\(=2\left(m^2-10m+25\right)-36=2\left(m-5\right)^2-36\ge-36\) \(\forall m\)