Giúp vs

a) \(\left(\right. x + y \left.\right)^{3} - \left(\right. x + y \left.\right) \left(\right. x^{2} - x y + y^{2} \left.\right) = 3 x y \left(\right. x + y \left.\right)\)

Giải:

Bắt đầu với vế trái của phương trình:

\(\left(\right. x + y \left.\right)^{3} - \left(\right. x + y \left.\right) \left(\right. x^{2} - x y + y^{2} \left.\right)\)

Bước 1: Mở rộng \(\left(\right. x + y \left.\right)^{3}\):

\(\left(\right. x + y \left.\right)^{3} = x^{3} + 3 x^{2} y + 3 x y^{2} + y^{3}\)

Bước 2: Mở rộng \(\left(\right. x + y \left.\right) \left(\right. x^{2} - x y + y^{2} \left.\right)\):

\(\left(\right. x + y \left.\right) \left(\right. x^{2} - x y + y^{2} \left.\right) = x \left(\right. x^{2} - x y + y^{2} \left.\right) + y \left(\right. x^{2} - x y + y^{2} \left.\right)\)\(= x^{3} - x^{2} y + x y^{2} + y x^{2} - x y^{2} + y^{3}\)\(= x^{3} + y^{3} + \left(\right. y x^{2} - x^{2} y \left.\right) = x^{3} + y^{3}\)

Bước 3: Trừ các biểu thức:

\(\left(\right. x + y \left.\right)^{3} - \left(\right. x + y \left.\right) \left(\right. x^{2} - x y + y^{2} \left.\right) = \left(\right. x^{3} + 3 x^{2} y + 3 x y^{2} + y^{3} \left.\right) - \left(\right. x^{3} + y^{3} \left.\right)\)\(= 3 x^{2} y + 3 x y^{2}\)\(= 3 x y \left(\right. x + y \left.\right)\)

Vậy, phương trình đã đúng:

\(\left(\right. x + y \left.\right)^{3} - \left(\right. x + y \left.\right) \left(\right. x^{2} - x y + y^{2} \left.\right) = 3 x y \left(\right. x + y \left.\right)\)

b) \(B = \left(\right. 3 x + 2 \left.\right) \left(\right. 9 x^{2} - 6 x + 4 \left.\right) - 3 \left(\right. 9 x^{3} - 2 \left.\right)\)

Giải:

Bước 1: Mở rộng \(\left(\right. 3 x + 2 \left.\right) \left(\right. 9 x^{2} - 6 x + 4 \left.\right)\):

\(\left(\right. 3 x + 2 \left.\right) \left(\right. 9 x^{2} - 6 x + 4 \left.\right) = 3 x \left(\right. 9 x^{2} - 6 x + 4 \left.\right) + 2 \left(\right. 9 x^{2} - 6 x + 4 \left.\right)\)\(= 27 x^{3} - 18 x^{2} + 12 x + 18 x^{2} - 12 x + 8\)\(= 27 x^{3} + 8\)

Bước 2: Mở rộng \(3 \left(\right. 9 x^{3} - 2 \left.\right)\):

\(3 \left(\right. 9 x^{3} - 2 \left.\right) = 27 x^{3} - 6\)

Bước 3: Trừ hai biểu thức:

\(B = \left(\right. 27 x^{3} + 8 \left.\right) - \left(\right. 27 x^{3} - 6 \left.\right) = 8 + 6 = 14\)

Vậy, \(B = 14\).

c) \(C = \left(\right. x - 2 \left.\right) \left(\right. x^{2} - 2 x + 4 \left.\right) - \left(\right. x^{3} - 7 \left.\right)\)

Giải:

Bước 1: Mở rộng \(\left(\right. x - 2 \left.\right) \left(\right. x^{2} - 2 x + 4 \left.\right)\):

\(\left(\right. x - 2 \left.\right) \left(\right. x^{2} - 2 x + 4 \left.\right) = x \left(\right. x^{2} - 2 x + 4 \left.\right) - 2 \left(\right. x^{2} - 2 x + 4 \left.\right)\)\(= x^{3} - 2 x^{2} + 4 x - 2 x^{2} + 4 x - 8\)\(= x^{3} - 4 x^{2} + 8 x - 8\)

Bước 2: Trừ biểu thức \(x^{3} - 7\):

\(C = \left(\right. x^{3} - 4 x^{2} + 8 x - 8 \left.\right) - \left(\right. x^{3} - 7 \left.\right)\)\(C = x^{3} - 4 x^{2} + 8 x - 8 - x^{3} + 7\)\(C = - 4 x^{2} + 8 x - 1\)

Vậy, \(C = - 4 x^{2} + 8 x - 1\).

d) \(D = \left(\right. x + 1 \left.\right)^{3} - \left(\right. x - 1 \left.\right) \left(\right. x^{2} + x + 1 \left.\right) - 3 x \left(\right. x + 1 \left.\right)\)

Giải:

Bước 1: Mở rộng \(\left(\right. x + 1 \left.\right)^{3}\):

\(\left(\right. x + 1 \left.\right)^{3} = x^{3} + 3 x^{2} + 3 x + 1\)

Bước 2: Mở rộng \(\left(\right. x - 1 \left.\right) \left(\right. x^{2} + x + 1 \left.\right)\):

\(\left(\right. x - 1 \left.\right) \left(\right. x^{2} + x + 1 \left.\right) = x \left(\right. x^{2} + x + 1 \left.\right) - 1 \left(\right. x^{2} + x + 1 \left.\right)\)\(= x^{3} + x^{2} + x - x^{2} - x - 1\)\(= x^{3} - 1\)

Bước 3: Mở rộng \(3 x \left(\right. x + 1 \left.\right)\):

\(3 x \left(\right. x + 1 \left.\right) = 3 x^{2} + 3 x\)

Bước 4: Trừ các biểu thức:

\(D = \left(\right. x^{3} + 3 x^{2} + 3 x + 1 \left.\right) - \left(\right. x^{3} - 1 \left.\right) - \left(\right. 3 x^{2} + 3 x \left.\right)\)\(D = x^{3} + 3 x^{2} + 3 x + 1 - x^{3} + 1 - 3 x^{2} - 3 x\)\(D = 2\)

Vậy, \(D = 2\).

e) \(E = 3 \left(\right. x - 1 \left.\right) \left(\right. x^{2} + x + 1 \left.\right) + x \left(\right. x + 1 \left.\right) - x \left(\right. x^{2} + x + 1 \left.\right)\)

Giải:

Bước 1: Mở rộng \(3 \left(\right. x - 1 \left.\right) \left(\right. x^{2} + x + 1 \left.\right)\):

\(3 \left(\right. x - 1 \left.\right) \left(\right. x^{2} + x + 1 \left.\right) = 3 \left(\right. x \left(\right. x^{2} + x + 1 \left.\right) - \left(\right. x^{2} + x + 1 \left.\right) \left.\right)\)\(= 3 \left(\right. x^{3} + x^{2} + x - x^{2} - x - 1 \left.\right) = 3 \left(\right. x^{3} - 1 \left.\right)\)\(= 3 x^{3} - 3\)

Bước 2: Mở rộng \(x \left(\right. x + 1 \left.\right)\):

\(x \left(\right. x + 1 \left.\right) = x^{2} + x\)

Bước 3: Mở rộng \(x \left(\right. x^{2} + x + 1 \left.\right)\):

\(x \left(\right. x^{2} + x + 1 \left.\right) = x^{3} + x^{2} + x\)

Bước 4: Trừ các biểu thức:

\(E = \left(\right. 3 x^{3} - 3 \left.\right) + \left(\right. x^{2} + x \left.\right) - \left(\right. x^{3} + x^{2} + x \left.\right)\)\(E = 3 x^{3} - 3 + x^{2} + x - x^{3} - x^{2} - x\)\(E = 2 x^{3} - 3\)

Vậy, \(E = 2 x^{3} - 3\).

g) \(9 x \left(\right. x + 1 \left.\right)^{3} + \left(\right. x - 1 \left.\right)^{3} = 2 x^{3}\)

Giải:

Mở rộng biểu thức và kiểm tra tính đúng đắn:

\(9 x \left(\right. x + 1 \left.\right)^{3} = 9 x \left(\right. x^{3} + 3 x^{2} + 3 x + 1 \left.\right) = 9 x^{4} + 27 x^{3} + 27 x^{2} + 9 x\)\(\left(\right. x - 1 \left.\right)^{3} = x^{3} - 3 x^{2} + 3 x - 1\)

Cộng cả hai biểu thức:

\(9 x \left(\right. x + 1 \left.\right)^{3} + \left(\right. x - 1 \left.\right)^{3} = 9 x^{4} + 27 x^{3} + 27 x^{2} + 9 x + x^{3} - 3 x^{2} + 3 x - 1\)\(= 9 x^{4} + 28 x^{3} + 24 x^{2} + 12 x - 1\)

So với \(2 x^{3}\), ta thấy biểu thức không đúng. Có thể bài toán có lỗi. Nếu có sự nhầm lẫn, bạn có thể điều chỉnh lại nhé!

h) \(\left(\right. x + 3 \left.\right) \left(\right. x^{2} - 3 x + 9 \left.\right) = x \left(\right. x^{2} - 3 x + 9 \left.\right) = x \left(\right. x^{2} + 4 \left.\right) - 1\)

Giúp mình nhé mọi người !

\(1.\)

\(a.\)

\(\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2}{x^2+3}+\dfrac{1}{x+1}\)

\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2\left(x^2-1\right)}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{1\left(x-1\right)\left(x^2+3\right)}{\left(x^2-1\right)\left(x^2+3\right)}\)

\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2x^2-2}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{x^3-x^2+3x-3}{\left(x^2-1\right)\left(x^2+3\right)}\)

\(=\dfrac{8+2x^2-2+x^3-x^2+3x-3}{\left(x^2+3\right)\left(x^2-1\right)}\)

\(=\dfrac{x^3+x^2+3x+3}{\left(x^2+3\right)\left(x^2-1\right)}\)

\(=\dfrac{x^2\left(x+1\right)+3\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)

\(=\dfrac{\left(x^2+3\right)\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)

\(=x-1\)

\(b.\)

\(\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{x^2-y^2}\)

\(=\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{\left(x-y\right)\left(x+y\right)}\)

\(=\dfrac{\left(x+y\right)^2}{2\left(x^2-y^2\right)}-\dfrac{\left(x-y\right)^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)

\(=\dfrac{x^2+2xy+y^2}{2\left(x^2-y^2\right)}-\dfrac{x^2-2xy+y^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)

\(=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+4y^2}{2\left(x^2-y^2\right)}\)

\(=\dfrac{4xy+4y^2}{2\left(x^2-y^2\right)}\)

\(=\dfrac{4y\left(x+y\right)}{2\left(x^2-y^2\right)}\)

\(=\dfrac{2y}{\left(x-y\right)}\)

Tương tự các câu còn lại

a/\(\left(x-1\right)\left(x^2+x+1\right)=x^3+x^2+x-x^2-x-1=x^3-1\left(đpcm\right)\)

b/ \(\left(x^3+x^2y+xy^2+y^3\right)\left(x-y\right)=x^4-x^3y+x^3y-x^2y^2+x^2y^2-xy^3+xy^3-y^4=x^4-y^4\left(đpcm\right)\)

c/ \(\left(x+y+z\right)^2=\left(x+y+z\right)\left(x+y+z\right)=x^2+xy+xz+y^2+xy+yz+z^2+zx+yz=x^2+y^2+z^2+2xy+2yz+2zx\left(đpcm\right)\)

d/ \(\left(x+y+z\right)^3=\left(x+y\right)^3+3\left(x+y\right)^2z+3z^2\left(x+y\right)+z^3\)

\(=\left(x+y\right)^3+3z\left(x^2+2xy+y^2\right)+3z^2\left(x+y\right)+z^3\)

\(=x^3+3x^2y+3xy^2+y^3+3x^2z+6xyz+3y^2z+3z^2x+3yz^2+z^3\)

\(=x^3+y^3+z^3+3xyz+3x^2y+3xy^2+3x^2z+3y^2z+3y^2x+3yz^2+3xyz\)

\(=x^3+y^3+z^3+\left(x+z\right)\left(3xy+3xz+3y^2+3yz\right)\)

\(=x^3+y^3+z^3+\left(x+z\right)\left[3x\left(y+z\right)+3y\left(y+z\right)\right]\)

\(=x^3+y^3+z^3+\left(x+z\right)\left(y+z\right)\left(3x+3y\right)\)

\(=x^3+y^3+z^3+3\left(x+y\right)\left(y+z\right)\left(z+x\right)\) (đpcm)

a, Xét vế trái ta có:

(x-1)(x^2+ x+1)=x^3+ x^2+ x- x^2- x-1

=x^3+ (x^2- x^2)+(x-x)-1

=x^3-1

Vậy...

b,Xét vế trái ta có:(x^3+ x^2y+ xy^2+ y^3)(x-y)

=x^4- x^3y+ x^3y- x^2- y^2+ x^2y^2- xy^3+ xy^3- y^4

=x^4-y^4

Vậy ........

c, Xét vế trái ta có:

(x+y+z)^2=(x+y+z)(x+y+z)

=x^2+ xy+ xz+ yx+y^2+ yz+ zx+ zy+ z^2

=x^2+ y^2+ z^2+ 2xy+ 2xz+ 2yz

Vậy...............

d, Xé vế trái ta có:

(x+y+x)^3=(x+y+z)(x+y+z)(x+y+z)(x+y+z)

=(x^2+y^2+z^2+2xy+2xz+2yz)(x+y+z)

=x^3+ xy^2+ xz^2+ 2x^2y+ 2xyz+ 2x^2z+ x^2y+ y^3+ yz^2+2xy^2+ 2y^2z+z^3+ 2xyz+ x^2z+ y^2z+2xyz+ 2yz^2+ 2xz^2

=x^3+ 3xy^2+ 6xy+ 3x^2y+3xz^2+ 3x^2z+ 3yz^2+ y^3z^3 (1)

Xét vế phải ta có:x^3+ y^3+ z^3+ 3(x+y)(x+y)(y+z)

=x^3+ y^3+ z^3+ 3(xy+ xz+ y^2+ yz)(z+x)

=x^3+ y^3+ z^3+ 3(xyz+ xz^2+ y^2z+ yz^2+ x^2y+ x^2z+ xy^2+xyz)

=x^2+ y^3+ z^3 +3(2xyz+ xz^2+ y^2z+ yz^2+x^2y+x^2z+ xy^2)

=x^3+ y^3+ z^3+6xyz+ 3xz^2+ 3y^2z+3yz^2+ 3x^2y+3x^2z+3xy^2(2)

Từ (1) và (2)=>.......

1.a (3x-2y)²= (3x)² - 2. 3x . 2y - (2y)² = 9x²  - 12xy - 4y²

2.b (2x - 1/2)² = (2x)² - 2.2x.1/2 - (1/2)²= 4x² - 2 - 1/4

3.c (x/2 - y) (x/2+y)= (x/2)² - (y)² = x/4 - y² 

Bài 1 :

 \(\left(3x-2y\right)^2=9x^2-12xy+4y^2\)

\(\left(2x-\frac{1}{2}\right)^2=4x^2-4x+\frac{1}{4}\)

\(\left(\frac{x}{2}-y\right)\left(\frac{x}{2}+y\right)=\frac{x^2}{4}-y^2\)

\(\left(x+\frac{1}{3}\right)^3=x^3+x^2+\frac{1}{3}x+\frac{1}{27}\)

\(\left(x-2\right)\left(x^2+2x+2^2\right)=x^3-8\)

dài lắm bạn ạ mk đang đau tay

lam giup minh di ma huheo

Vậy (x^4 - x^3 - 3x^2 + x + 2) = (x^2 - x - 1)(x^2 - 1) + 1

\(P=\left(x-y\right)^2+\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)-4x^2=\left(x-y-x-y\right)^2-\left(2x\right)^2=\left(-2y\right)^2-\left(2x\right)^2\)

\(=\left(2y-2x\right)\left(2y+2x\right)=2\left(y-x\right)2\left(y+x\right)=4\left(x+y\right)\left(y-x\right)\)

\(x^3-x^2y+3x-3y=x^2\left(x-y\right)+3\left(x-y\right)=\left(x-y\right)\left(x^2+3\right)\)

\(x^3-2x^2-4xy^2+x=x\left(x^2-2x+1-4y^2\right)=x\left[\left(x-1\right)^2-\left(2y\right)^2\right]=x\left(x+2y-1\right)\left(x-2y-1\right)\)

\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-8=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-8\)

Đặt \(x^2+7x+10=t\), ta có:

\(t\left(t+2\right)-8=t^2+2t-8=t^2-2t+4t-8=t\left(t-2\right)+4\left(t-2\right)=\left(t-2\right)\left(t+4\right)\)

\(=\left(x^2+7x+10+4\right)\left(x^2+7x+10-2\right)=\left(x^2+7x+14\right)\left(x^2+7x-8\right)\)

1,Thực hiện phép tính :

a, (x + 2)⁹ : (x + 2)⁶

=(x+2)^9-6

=(x+2)³

b, (x - y) ⁴ : (x - 2)³

=(x-y)^4-3

=x-y

c, ( x²+ 2x + 4)⁵ : (x² + 2x + 4)

=(x²+2x+4)^5-1

=(x²+2x+4)⁴

d, 2(x² + 1)³ : 1/3(x² + 1)

=(2÷1/3).[(x²+1)³÷(x²+1)]

=6(x²+1)²

e, 5 (x - y)⁵ : 5/6 (x - y)²

=(5÷5/6).[(x-y)⁵÷(x-y)²]

=6(x-y)⁾³

\(\left\{{}\begin{matrix}x+y=13\\xy=22\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2=x^2+2xy+y^2=169\\4xy=88\end{matrix}\right.\Leftrightarrow x^2+2xy+y^2-4xy=81=\left(\pm9\right)^2\) \(+,x-y=9\Rightarrow\left\{{}\begin{matrix}x+y=13\\x-y=9\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=11\\y=2\end{matrix}\right.\)

\(+,x-y=-9\Rightarrow\left\{{}\begin{matrix}x+y=13\\x-y=-9\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\y=11\end{matrix}\right.\)

\(\Rightarrow x^2+y^2=11^2+2^2=125;x^3+y^3=11^3+2^3=1339;x^4-y^4=\left(x^2+y^2\right)\left(x^2-y^2\right)=\pm\left(11^2+2^2\right)\left(11^2-2^2\right)=\pm14625;x^7+y^7=11^7+2^7=19487299;x-y=\pm\left(11-2\right)=\pm9\)

\(a+b+c=0\Rightarrow\left(a+b+c\right)^2=\left(a^2+b^2+c^2\right)+2\left(ab+bc+ca\right)=0\Rightarrow ab+bc+ca=-\frac{1}{2}\Rightarrow\left(ab+bc+ca\right)^2=a^2b^2+b^2c^2+c^2a^2+2\left(ab^2c+abc^2+a^2bc\right)=\frac{1}{4}\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+\left(a+b+c\right)abc=\frac{1}{4}\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+0=\frac{1}{4}\Leftrightarrow a^2b^2+b^2c^2+c^2a^2=\frac{1}{4}\Leftrightarrow2\left(a^2b^2+b^2c^2+c^2a^2\right)=\frac{1}{2};\left(a^2+b^2+c^2\right)^2=a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=1^2=1\)

\(\Rightarrow\left(a^4+b^4+c^4\right)+\frac{1}{2}=1\Rightarrow\left(a^4+b^4+c^4\right)=\frac{1}{2}\Leftrightarrow A=\frac{1}{2}\)

đề

Tìm x,y,z biết