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\(\left(x+\frac{4}{3}y^2\right)^2=x^2+\frac{8xy^2}{3}+\frac{16y^4}{9}\)
\(\left(2x^2+\frac{5}{3}y\right)^2=4x^4+\frac{20x^2y}{3}+\frac{25y^2}{9}\)
\(\left(x+\frac{4}{3}y^2\right)^2=x^2+2\cdot x\cdot\frac{4}{3}y^2+\left(\frac{4}{3}y^2\right)^2=x^2+\frac{8}{3}xy^2+\frac{16}{9}y^4\)
\(\left(2x^2+\frac{5}{3}y\right)^2=\left(2x^2\right)^2+2\cdot2x^2\cdot\frac{5}{3}y+\left(\frac{5}{3}y\right)^2=4x^4+\frac{20}{3}x^2y+\frac{25}{9}y^2\)
Giải:
a) \(\left(x^2+x-1\right)^2-\left(x^2+2x+3\right)^2\)
\(=\left[\left(x^2+x-1\right)-\left(x^2+2x+3\right)\right]\left[\left(x^2+x-1\right)+\left(x^2+2x+3\right)\right]\)
\(=\left(x^2+x-1-x^2-2x-3\right)\left(x^2+x-1+x^2+2x+3\right)\)
\(=\left(-x-4\right)\left(2x^2+3x+2\right)\)
Vậy ...
b) \(-16+\left(x-3\right)^2\)
\(=\left(x-3\right)^2-16\)
\(=\left(x-3\right)^2-4^2\)
\(=\left(x-3-4\right)\left(x-3+4\right)\)
\(=\left(x-7\right)\left(x+1\right)\)
Vậy ...
c) \(64+16y+y^2\)
\(=8^2+2.8.y+y^2\)
\(=\left(8+y\right)^2\)
Vậy ...
a,16x2-9
=42x2-32
=(4x-3)(4x+3) HĐT thứ 3
b,9a2-25b2
=32a2-52b2
=(3a-5b)(3a+5b) HĐT thứ 3
c,81-y4
=32.32-y2.y2
=(32-y2)
=(3-y)(3+y) HĐT thứ 3
d,(2x+y)2-1
=(2x+y-1)(2x+y-1) HĐT thứ 3
e,(x+y+z)2-(x-y-z)2
cái này là HĐT thứ 8 mở rộng bạn lên mạng tìm nha
a) \(16x^2-9=\left(4x\right)^2-3^2=\left(4x-3\right).\left(4x+3\right)\)
b) \(9a^2-25b^4=\left(3a\right)^2-\left(5b^2\right)^2=\left(3a-5b^2\right).\left(3a+5b^2\right)\)
c) \(81-y^4=9^2-\left(y^2\right)^2=\left(9-y^2\right).\left(9+y^2\right)\)
d)\(\left(2x+y\right)^2-1=\left(2x+y\right)^2-1^2=\left(2x+y-1\right).\left(2x+y+1\right)\)
đề dài v~
1.
a) \(f\left(x\right)=5x^2-2x+1\)
\(5f\left(x\right)=25x^2-10x+5\)
\(5f\left(x\right)=\left(25x^2-10x+1\right)+4\)
\(5f\left(x\right)=\left(5x-1\right)^2+4\)
Mà \(\left(5x-1\right)^2\ge0\)
\(\Rightarrow5f\left(x\right)\ge4\)
\(\Leftrightarrow f\left(x\right)\ge\frac{4}{5}\)
Dấu " = " xảy ra khi :
\(5x-1=0\Leftrightarrow x=\frac{1}{5}\)
Vậy ....
b) \(P\left(x\right)=3x^2+x+7\)
\(3P\left(x\right)=9x^2+3x+21\)
\(3P\left(x\right)=\left(9x^2+3x+\frac{1}{4}\right)+\frac{83}{4}\)
\(3P\left(x\right)=\left(3x+\frac{1}{2}\right)^2+\frac{83}{4}\)
Mà \(\left(3x+\frac{1}{2}\right)^2\ge0\)
\(\Rightarrow3P\left(x\right)\ge\frac{83}{4}\)
\(\Leftrightarrow P\left(x\right)\ge\frac{83}{12}\)
Dấu "=" xảy ra khi :
\(3x+\frac{1}{2}=0\Leftrightarrow x=-\frac{1}{6}\)
Vậy ...
c) \(Q\left(x\right)=5x^2-3x-3\)
\(5Q\left(x\right)=25x^2-15x-15\)
\(\Leftrightarrow5Q\left(x\right)=\left(25x^2-15x+\frac{9}{4}\right)-\frac{69}{4}\)
\(\Leftrightarrow5Q\left(x\right)=\left(5x-\frac{3}{2}\right)^2-\frac{69}{4}\)
Mà \(\left(5x-\frac{3}{2}\right)^2\ge0\)
\(\Rightarrow5Q\left(x\right)\ge\frac{-69}{4}\)
\(\Leftrightarrow Q\left(x\right)\ge-\frac{69}{20}\)
Dấu "=" xảy ra khi :
\(5x-\frac{3}{2}=0\Leftrightarrow x=0,3\)
Vậy ...
2.
a) \(f\left(x\right)=-3x^2+x-2\)
\(-3f\left(x\right)=9x^2-3x+6\)
\(-3f\left(x\right)=\left(9x^2-3x+\frac{1}{4}\right)+\frac{23}{4}\)
\(-3f\left(x\right)=\left(3x-\frac{1}{2}\right)^2+\frac{23}{4}\)
Mà \(\left(3x-\frac{1}{2}\right)^2\ge0\)
\(\Rightarrow-3f\left(x\right)\ge\frac{23}{4}\)
\(\Leftrightarrow f\left(x\right)\le\frac{23}{12}\)
Dấu "=" xảy ra khi :
\(3x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{6}\)
Vậy ...
b) \(P\left(x\right)=-x^2-7x+1\)
\(-P\left(x\right)=x^2+7x-1\)
\(-P\left(x\right)=\left(x^2+7x+\frac{49}{4}\right)-\frac{53}{4}\)
\(-P\left(x\right)=\left(x+\frac{7}{2}\right)^2-\frac{53}{4}\)
Mà \(\left(x+\frac{7}{2}\right)^2\ge0\)
\(\Rightarrow-P\left(x\right)\ge-\frac{53}{4}\)
\(\Leftrightarrow P\left(x\right)\le\frac{53}{4}\)
Dấu "=" xảy ra khi :
\(x+\frac{7}{2}=0\Leftrightarrow x=-\frac{7}{2}\)
Vậy ...
c) \(Q\left(x\right)=-2x^2+x-8\)
\(-2Q\left(x\right)=4x^2-2x+16\)
\(-2Q\left(x\right)=\left(4x^2-2x+\frac{1}{4}\right)+\frac{63}{4}\)
\(-2Q\left(x\right)=\left(2x-\frac{1}{2}\right)^2+\frac{63}{4}\)
Mà : \(\left(2x-\frac{1}{2}\right)^2\ge0\)
\(\Rightarrow-2Q\left(x\right)\ge\frac{63}{4}\)
\(\Leftrightarrow Q\left(x\right)\le-\frac{63}{8}\)
Dấu "=" xảy ra khi :
\(2x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{4}\)
Vậy ...
Vì cả 2 số hạng đều là số chính phương, ta phân tích nhân tử bằng cách sử dụng công thức hiệu của 2 bình phương:\(a^2-b^2=\left(a+b\right)\left(a-b\right)\) trong đó: \(a=x^2+x-1\)và \(b=x^2+2x+3\)
\(\Rightarrow\left(2x^2+3x+2\right)\left(x+4\right)\)
1)
a) \(x^2+12x+36=\left(x+6\right)^2\)
b) \(x^2-x+\dfrac{1}{4}=\left(x-\dfrac{1}{2}\right)^2\)
Tick nha
3)
a)\(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)
\(\Leftrightarrow x^3+8-x^3-2x=15\)
\(\Leftrightarrow-2x=15-8\)
\(\Leftrightarrow-2x=7\)
\(\Rightarrow x=\dfrac{-7}{2}\)
b) \(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2\right)-5x+1=28\)
\(\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3-10x^2+2x+4x^2-5x+1=28\)
\(\Leftrightarrow0-3x^2+23x+28=28\)
\(\Leftrightarrow-3x^2+23x=0\)
\(\Leftrightarrow-x\left(3x-23\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-x=0\\3x-23=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{23}{3}\end{matrix}\right.\)
c) \(\left(x^2-1\right)^3-\left(x^4+x^2+1\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow x^6-3x^4+3x^2-1-x^6-2x^4-2x^2-1=0\)
\(\Leftrightarrow-5x^4+x^2-2=0\)
Đặt \(-5t^2+t-2=0\)
\(\Delta=1^2-4\left(-5\right)\left(-2\right)=-39< 0\)
\(\Rightarrow PTVN\)
Bài 2 :
a ) \(A=\left(a+b+c\right)^2+a^2+b^2+c^2\)
\(A=a^2+b^2+c^2+2ab+2ac+2bc+a^2+b^2+c^2\)
\(A=\left(a^2+2ab+b^2\right)+\left(a^2+2ac+c^2\right)+\left(b^2+2bc+c^2\right)\)
\(A=\left(a+b\right)^2+\left(a+c\right)^2+\left(b+c\right)^2\)
\(\left(\dfrac{1}{2}+x\right)^2=\dfrac{1}{4}+x+x^2\)
\(\left(2x+1\right)^2=4x^2+4x+1\)
1. (1/2 +x)2= (1/2)2 + x +x2 = 1/4 +x +x2
(2x+1)2 = 4x2 +4x +1
chúc bạn học tốt