c)2a2b + 4ab2- a2c + ac2- 4b2c + 2bc2- 4abc;
d) y(x - 2z)2 + 8xyz + x(y - 2z)2- 2z(x + y)2
e) x5- 5x3 + 4x;
Những câu hỏi liên quan
1). x2y2(y-x)+y2z2(z-y)-z2x2(z-x)
2)xyz-(xy+yz+xz)+(x+y+z)-1
3)yz(y+z)+xz(z-x)-xy(x+y)
4)2a2b+4ab2-a2c+ac2-4b2c+2bc2-4abc
5)y(x-2z)2+8xyz+x(y-2z)2-2z(x+y)2
6)8x3(y+z)-y3(z+2x)-z3(2x-y)
7) (x2+y2)3+(z2-x2)3-(y2+z2)3
1). x2y2(y-x)+y2z2(z-y)-z2x2(z-x)
2)xyz-(xy+yz+xz)+(x+y+z)-1
3)yz(y+z)+xz(z-x)-xy(x+y)
4)2a2b+4ab2-a2c+ac2-4b2c+2bc2-4abc
5)y(x-2z)2+8xyz+x(y-2z)2-2z(x+y)2
6)8x3(y+z)-y3(z+2x)-z3(2x-y)
7) (x2+y2)3+(z2-x2)3-(y2+z2)3
bn gõ bài trong công thức trực quan ik, khó nhìn lắm, ko làm đc
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1) \(x^2y^2\left(y-x\right)+y^2z^2\left(z-y\right)-z^2x^2\left(z-x\right)\)
\(=x^2y^3-x^3y^2+y^2z^3-y^3z^2-z^2x^2\left(z-x\right)\)
\(=\left(y^2z^3-x^3y^2\right)-\left(y^3z^2-x^2y^3\right)-z^2x^2\left(z-x\right)\)
\(=y^2\left(z^3-x^3\right)-y^3\left(z^2-x^2\right)-z^2x^2\left(z-x\right)\)
\(=y^2\left(z-x\right)\left(z^2+zx+x^2\right)-y^3\left(z-x\right)\left(z+x\right)-z^2x^2\left(z-x\right)\)
\(=\left(z-x\right)\left[y^2\left(z^2+zx+x^2\right)-y^3\left(z+x\right)-z^2x^2\right]\)
\(=\left(z-x\right)\left[\left(y^2z^2+xy^2z+x^2y^2\right)-\left(y^3z+xy^3\right)-z^2x^2\right]\)
\(=\left(z-x\right)\left(y^2z^2+xy^2z+x^2y^2-y^3z-xy^3-z^2x^2\right)\)
\(=\left(z-x\right)\left[\left(y^2z^2-y^3z\right)-\left(x^2z^2-x^2y^2\right)+\left(xy^2z-xy^3\right)\right]\)
\(=\left(z-x\right)\left[y^2z\left(z-y\right)-x^2\left(z^2-y^2\right)+xy^2\left(z-y\right)\right]\)
\(=\left(z-x\right)\left[y^2z\left(z-y\right)-x^2\left(z-y\right)\left(z+y\right)+xy^2\left(z-y\right)\right]\)
\(=\left(z-x\right)\left(z-y\right)\left[y^2z-x^2\left(z+y\right)+xy^2\right]\)
\(=\left(z-x\right)\left(z-y\right)\left(y^2z-x^2z-x^2y+xy^2\right)\)
\(=\left(z-x\right)\left(z-y\right)\left[\left(y^2z-x^2z\right)-\left(x^2y-xy^2\right)\right]\)
\(=\left(z-x\right)\left(z-y\right)\left[z\left(y^2-x^2\right)-xy\left(x-y\right)\right]\)
\(=\left(z-x\right)\left(z-y\right)\left[z\left(y-x\right)\left(y+x\right)+xy\left(y-x\right)\right]\)
\(=\left(z-x\right)\left(z-y\right)\left(y-x\right)\left[z\left(y+x\right)+xy\right]\)
\(=\left(z-x\right)\left(z-y\right)\left(y-x\right)\left(yz+xz+xy\right)\)
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2) \(xyz-\left(xy+yz+xz\right)+\left(x+y+z\right)-1\)
\(=xyz-xy-yz-xz+x+y+z-1\)
\(=\left(xyz-xy\right)-\left(yz-y\right)-\left(xz-x\right)+\left(z-1\right)\)
\(=xy\left(z-1\right)-y\left(z-1\right)-x\left(z-1\right)+\left(z-1\right)\)
\(=\left(z-1\right)\left(xy-y-x+1\right)\)
\(=\left(z-1\right)\left[\left(xy-y\right)-\left(x-1\right)\right]\)
\(=\left(z-1\right)\left[y\left(x-1\right)-\left(x-1\right)\right]\)
\(=\left(z-1\right)\left(x-1\right)\left(y-1\right)\)
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Cho phương trình bậc hai: ax2+bx+c=0 có hai nghiệm x1,x2thỏa mãn ax1+bx2+c=0. Tính M=a2c+ac2+b3-3abc+2018
phân tích đa thức thành nhân tử: a2b+a2c+ab2+ac2+c2b+cb2+2abc
Làm tính chia: 5 a 3 b : - 2 a 2 b
5 a 3 b : - 2 a 2 b = 5 : - 2 . a 3 : a 2 b : b = - 5 / 2 . a
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a) 25x2 - 20xy + 4y
b) 1/36a2 - 1/4b2
c) 0.125 (a+2)3 - 1
d) x6 - 1
\(a,Sửa:25x^2-20xy+4y^2=\left(5x-2y\right)^2\\ b,=\dfrac{1}{4}\left(\dfrac{1}{9}a^2-b^2\right)=\dfrac{1}{4}\left(\dfrac{1}{3}a-b\right)\left(\dfrac{1}{3}a+b\right)\\ c,=\dfrac{1}{8}\left(a+2\right)^3-1=\left[\dfrac{1}{2}\left(a+2\right)\right]^3-1=\left[\dfrac{1}{2}a+1\right]^3-1\\ =\left(\dfrac{1}{2}a+1-1\right)\left(\dfrac{1}{4}a^2+a+1+\dfrac{1}{2}a+1+1\right)\\ =\dfrac{1}{2}a\left(\dfrac{1}{4}a^2+\dfrac{3}{2}a+3\right)\\ d,=\left(x^3-1\right)\left(x^3+1\right)\\ =\left(x-1\right)\left(x+1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)\)
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a2−4b2−a3+ 4a2b−4ab2 phân tích đa thức thành nhân tử
Cho ba số dương a, b, c. Chứng minh bất đẳng thức: \(ab+bc+ca\ge\dfrac{4abc}{2a+b+c}+\dfrac{4abc}{2b+c+a}+\dfrac{4abc}{2c+a+b}\)
\(BDT\Leftrightarrow\dfrac{1}{4a}+\dfrac{1}{4b}+\dfrac{1}{4c}\ge\dfrac{1}{2a+b+c}+\dfrac{1}{2b+c+a}+\dfrac{1}{2c+a+b}\)
Áp dụng BĐT \(\dfrac{1}{nht}+\dfrac{1}{is}+\dfrac{1}{the}+\dfrac{1}{best}\ge\dfrac{16}{nht+is+the+best}\):
\(\dfrac{1}{2a+b+c}\le\dfrac{1}{16}\left(\dfrac{1}{a}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
Tương tự cho 2 BĐT còn lại rồi cộng theo vế:
\(VP\le\dfrac{4}{16}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{4a}+\dfrac{1}{4b}+\dfrac{1}{4c}\)
\("="\Leftrightarrow a=b=c\)
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10. Cho 
vuông tại B chọn câu đúng
a.BC2 = AB2 + AC2 b. AB2 = AC2 + BC2 c. AC2 = BC2 + AB2
