Câu 1:

(2\(x\) - 8)\(^2\)

= (2\(x)^2\) - 2.2\(x\) .8 + 8\(^2\)

= 4\(x^2\) - (2.2.8)\(x\) + 64

= 4\(x^2\) - 4.8\(x\) + 64

= 4\(x^2\) - 32\(x\) + 64

Câu 2:

(\(x-8)^2\)

= \(x^2\) - 2.\(x.8\) + 8\(^2\)

= \(x^2\) - 2.8.\(x\) + 64

= \(x^2\) - 16\(x\) + 64

Hi

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

\(4x=x+2\)

\(3x=2\)

\(x=\frac23\)

tick giùm mình nha bạn

C =(a - b - c)\(^2\) - a\(^2\) - b\(^2\) - c\(^2\)

C = (a\(^{}\) - b)\(^2\) - 2(a -b)c + c\(^2\) - a\(^2\) - b\(^2\) - \(c^2\)

C = a\(^2\) - 2ab + b\(^2\) - 2ac + 2bc + c\(^2\) - \(a^2\) - \(b^2-c^2\)

C = (a\(^2\) - a\(^2\))+(\(b^2\) - b\(^2\))+(c\(^2\) - \(c^2\))-2ab - 2ac + 2bc

C = 0 + 0 + 0 - 2ab - 2ac + 2bc

C = -2ab - 2ac + 2bc

x trái dấu với y

=>xy<0

=>\(-2abc^3\cdot3a^2b^3c^5<0\)

=>\(-6a^3b^4c^8<0\)

=>\(-6a^3<0\)

=>\(a^3>0\)

=>a>0

\(x=-2abc^3\)

\(y=3a^2b^3c^5\)

Ta có:

\(xy=-2abc^3.\left(3a^2b^3c^3\right)\)

\(xy=-2a^3b^4c^6\)

Do \(x\) và \(y\) trái dấu \(\rArr xy=-2a^3b^4c^6<0\)

Xét:

\(b^4\ge0\) (mọi \(b\))

\(c^6\ge0\) (mọi \(c\))

Để \(-2a^3b^4c^6<0\rArr a^3\) dương

\(\rArr a\) mang dấu dương

Ta gọi biểu thức là:

\(A = x^{3} + \left[\right. \left(\right. x^{2} - 2 x + 2 \left.\right)^{2} - x \left(\right. x^{3} + 8 x - 7 \left.\right) - 4 \left]\right.\)

Bước 1: Khai triển và rút gọn

Tính \(\left(\right. x^{2} - 2 x + 2 \left.\right)^{2}\):

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

Tính \(x \left(\right. x^{3} + 8 x - 7 \left.\right)\):

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

Thay vào biểu thức \(A\):

\(A = x^{3} + \left[\right. \left(\right. x^{4} - 4 x^{3} + 8 x^{2} - 8 x + 4 \left.\right) - \left(\right. x^{4} + 8 x^{2} - 7 x \left.\right) - 4 \left]\right.\)

Rút gọn:

\(A = x^{3} + \left[\right. x^{4} - 4 x^{3} + 8 x^{2} - 8 x + 4 - x^{4} - 8 x^{2} + 7 x - 4 \left]\right.\) \(A = x^{3} + \left(\right. - 4 x^{3} - x \left.\right)\) \(A = x^{3} - 4 x^{3} - x = - 3 x^{3} - x\)

Bước 2: Phân tích A

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

Bước 3: Chứng minh chia hết cho 6

-Với mọi \(x \in \mathbb{Z}\), thì:

-Nếu \(x\) chẵn → chia hết cho 2

-Nếu \(x\) bội của 3 → chia hết cho 3
→ Luôn có \(A\) chia hết cho 6 với mọi \(x \in \mathbb{Z}\)

Vậy biểu thức A chia hết cho 6.

Đặt \(A=x^3+\left\lbrack\left(x^2-2x+2\right)^2-x\left(x^3+8x-7\right)-4\right\rbrack\)

\(=x^3+\left\lbrack x^4+4x^2+4-4x^3+4x^2-8x-x\left(x^3+8x-7\right)-4\right\rbrack\)

\(=x^3+\left\lbrack x^4-4x^3+8x^2-8x-x^4-8x^2+7x\right\rbrack\)

\(=x^3+\left(-4x^3-x\right)=-3x^3-x\)

Khi x=1 thì \(A=-3\cdot1^3-1=-3-1=-4\) không chia hết cho 6

=>Đề sai rồi bạn

\(\left(2x-1\right)\left(4x^2+2x+1\right)-2x\left(2x-3\right)\left(2x+3\right)=81x^2\)

\(8x^3-1-2x\left(4x^2-9\right)=81x^2\)

\(8x^3-1-\left(8x^3-18x\right)=81x^2\)

\(8x^3-1-8x^3+18x=81x^2\)

\(18x-1=81x^2\)

\(81x^2-18x+1=0\)

\(\left(9x\right)^2-2\cdot9x+1=0\)

\(\left(9x-1\right)^2=0\)

\(9x-1=0\)

\(x=\frac19\)

Vậy \(x=\frac19\)

Ta có: \(\left(2x-1\right)\left(4x^2+2x+1\right)-2x\left(2x-3\right)\left(2x+3\right)=81x^2\)

=>\(8x^3-1-2x\left(4x^2-9\right)=81x^2\)

=>\(8x^3-1-8x^3+18x=81x^2\)

=>\(81x^2=18x-1\)

=>\(81x^2-18x+1=0\)

=>\(\left(9x-1\right)^2=0\)

=>9x-1=0

=>9x=1

=>\(x=\frac19\)

Sửa đề: \(2a^2+7ab+3b^2=0\)

=>\(2a^2+6ab+ab+3b^2=0\)

=>2a(a+3b)+b(a+3b)=0

=>(a+3b)(2a+b)=0

=>\(\left[\begin{array}{l}a+3b=0\\ 2a+b=0\end{array}\right.\Rightarrow\left[\begin{array}{l}a=-3b\\ b=-2a\end{array}\right.\)

TH1: a=-3b

\(\frac{8a-3b}{2a-b}-\frac{2a-5b}{2a+b}\)

\(=\frac{8\cdot\left(-3b\right)-3b}{2\left(-3b\right)-b}-\frac{2\cdot\left(-3b\right)-5b}{2\cdot\left(-3b\right)+b}=\frac{-24b-3b}{-6b-b}-\frac{-6b-5b}{-6b+b}\)

\(=\frac{-27}{-7}-\frac{-11}{-5}=\frac{27}{7}-\frac{11}{5}=\frac{135}{35}-\frac{77}{35}=\frac{58}{35}\)

TH2: b=-2a

\(\frac{8a-3b}{2a-b}-\frac{2a-5b}{2a+b}\)

\(=\frac{8a-3\cdot\left(-2a\right)}{2a-\left(-2a\right)}-\frac{2a-5\cdot\left(-2a\right)}{2a-2a}=\frac{14a}{4a}-\frac{12a}{0a}\)

=>Khi b=-2a thì biểu thức không có giá trị