(\(x+1\))² - (\(x+1\)) = 0

(\(x+1\))(\(x+1-1\)) =0 

(\(x+1\))\(x\)   = 0

\(\left[{}\begin{matrix}x+1=0\\x=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=-1\\x=0\end{matrix}\right.\)

Vậy \(x\in\){ -1; 0}

Giải bằng cách phân tích đa thức thành nhân tử nhé mn

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

\(\Rightarrow D=x^6-3x^4+3x^2-1-\left(x^6-1\right)\)

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

a) \(\left(2x-1\right)^3-4x^2\left(2x-3\right)=5\)

\(\Leftrightarrow8x^3-12x^2+6x-1-8x^3+12x^2=5\)

\(\Leftrightarrow6x-1=5\Leftrightarrow6x=6\Leftrightarrow x=1\)

b) \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x+1\right)^2=-10\)

\(\Leftrightarrow\left(x+1-x+1\right)\left[\left(x^2+2x+1+x^2-2x+1+\left(x^2-1\right)\right)\right]-6\left(x^2+2x+1\right)=-10\)

\(\Leftrightarrow2\left(3x^2+1\right)-6x^2-12x-6=-10\)

\(\Leftrightarrow6x^2+2-6x^2-12x-6=-10\)

\(\Leftrightarrow-12x-4=-10\Leftrightarrow12x=-6\Leftrightarrow x=\dfrac{1}{2}\)

\(B=\left(x-3\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+\left(3x-1\right)\left(3x+1\right)\)

\(B=x^3-9x^2+27x-27-\left(x^3-3x^2+9x+3x^2-9x+27\right)+\left(9x^2-1\right)\)

\(B=x^3-9x^2+27x-27-\left(x^3+27\right)+9x^2-1\)

\(B=x^3-9x^2+27x-27-x^3-27+9x^2-1\)

\(B=27x-55\)

giải cho tui cả phần c nx đi

Yêu cầu đề bài của bạn?

đề bài : Tính

Để chứng minh rằng √(a-b) và √(3a+3b+1) là các số chính phương, ta sẽ điều chỉnh phương trình ban đầu để tìm mối liên hệ giữa các biểu thức này. Phương trình ban đầu: 2^(2+a) = 3^(2+b) Ta có thể viết lại phương trình theo dạng: (2^2)^((1/2)+a/2) = (3^2)^((1/2)+b/2) Simplifying the exponents, we get: 4^(1/2)*4^(a/2) = 9^(1/2)*9^(b/2) Taking square roots of both sides, we have: √4*√(4^a) = √9*√(9^b) Simplifying further, we obtain: 22*(√(4^a)) = 32*(√(9^b)) Since (√x)^y is equal to x^(y/), we can rewrite the equation as follows: 22*(4^a)/ = 32*(9^b)/ Now let's examine the expressions inside the square roots: √(a-b) can be written as (√((22*(4^a))/ - (32*(9^b))/)) Similarly, √(3*a + 3*b + ) can be written as (√((22*(4^a))/ + (32*(9^b))/)) We can see that both expressions are in the form of a difference and sum of two squares. Therefore, it follows that both √(a-b) and √(3*a + 3*b + ) are perfect squares.

\(p=\left[\left(x+5\right).\left(x+11\right)\right].\left[\left(x+7\right).\left(x+9\right)\right]+16=\)

\(=\left(x^2+16x+55\right)\left(x^2+16x+63\right)+16=\)

\(=\left(x^2+16x\right)^2+118.\left(x^2+16x\right)+3481=\)

\(=\left(x^2+16x\right)^2+2.\left(x^2+16x\right).59+59^2=\)

\(=\left[\left(x^2+16x\right)+59\right]^2\) là một số chính phương

\(Q=\left(a^2b^2+a^2+b^2+1\right)\left(c^2+1\right)=\)

\(=a^2b^2c^2+a^2b^2+a^2c^2+a^2+b^2c^2+b^2+c^2+1=\)

\(=a^2b^2c^2+\left(a^2b^2+b^2c^2+a^2c^2\right)+\left(a^2+b^2+c^2\right)+1\) (1)

Ta có

\(\left(ab+bc+ac\right)^2=a^2b^2+b^2c^2+a^2c^2+2ab^2c+2abc^2+2a^2bc=\)

\(=a^2b^2+b^2c^2+a^2c^2+2abc\left(a+b+c\right)=1\)

\(\Rightarrow a^2b^2+b^2c^2+a^2c^2=1-2abc\left(a+b+c\right)\) (2)

Ta có

\(\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ac\right)=\)

\(=a^2+b^2+c^2+2\)

\(\Rightarrow a^2+b^2+c^2=\left(a+b+c\right)^2-2\) (3)

Thay (2) và (3) vào (1)

\(Q=a^2b^2c^2+1-2abc\left(a+b+c\right)+\left(a+b+c\right)^2-2+1=\)

\(=\left(abc\right)^2-2abc\left(a+b+c\right)+\left(a+b+c\right)^2=\)

\(=\left[abc-\left(a+b+c\right)\right]^2\)

\(n=a^2+b^2\)

\(\Rightarrow n^2=\left(a^2+b^2\right)^2-4a^2b^2+4a^2b^2=\)

\(=\left(a^2+b^2-2ab\right)\left(a^2+b^2+2ab\right)+\left(2ab\right)^2=\)

\(=\left(a-b\right)^2\left(a+b\right)^2+\left(2ab\right)^2=\)

\(=\left[\left(a-b\right)\left(a+b\right)\right]^2+\left(2ab\right)^2=\)

\(=\left(a^2-b^2\right)^2+\left(2ab\right)^2\)

\(A=\left(x^2+x+1\right)^2-4\left(x+2\right)^2+15\)

\(\Rightarrow A=\left(x^2+x+1\right)^2-\left[2\left(x+2\right)\right]^2+15\)

\(\Rightarrow A=\left(x^2+x+1+2x+2\right)\left(x^2+x+1-2x-2\right)+15\)

\(\Rightarrow A=\left(x^2+3x+3\right)\left(x^2-x-1\right)+15\)

\(\Rightarrow A=\left(x^2+3x+\dfrac{9}{4}-\dfrac{9}{4}+3\right)\left(x^2-x+\dfrac{1}{4}-\dfrac{1}{4}-1\right)+15\)

\(\Rightarrow A=\left[\left(x+\dfrac{3}{2}\right)^2+\dfrac{3}{4}\right]\left[\left(x-\dfrac{1}{2}\right)^2-\dfrac{5}{4}\right]+15\left(1\right)\)

Ta có : \(\left\{{}\begin{matrix}\left(x+\dfrac{3}{2}\right)^2\ge0,\forall x\\\left(x-\dfrac{1}{2}\right)^2\ge0,\forall x\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left(x+\dfrac{3}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4},\forall x\\\left(x-\dfrac{1}{2}\right)^2-\dfrac{5}{4}\ge-\dfrac{5}{4},\forall x\end{matrix}\right.\)

\(\left(1\right)\Rightarrow\left[{}\begin{matrix}A\ge\dfrac{3}{4}.\left[\left(-\dfrac{3}{2}-\dfrac{1}{2}\right)^2-\dfrac{5}{4}\right]+15\left(x=-\dfrac{3}{2}\right)\\A\ge\left[\left(\dfrac{1}{2}+\dfrac{3}{2}\right)^2+\dfrac{3}{4}\right].\left(-\dfrac{5}{4}\right)+15\left(x=\dfrac{1}{2}\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}A\ge\dfrac{3}{4}.\left[4-\dfrac{5}{4}\right]+15\left(x=-\dfrac{3}{2}\right)\\A\ge\left[4+\dfrac{3}{4}\right].\left(-\dfrac{5}{4}\right)+15\left(x=\dfrac{1}{2}\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}A\ge\dfrac{3}{4}.\dfrac{9}{4}+15\left(x=-\dfrac{3}{2}\right)\\A\ge\dfrac{19}{4}.\left(-\dfrac{5}{4}\right)+15\left(x=\dfrac{1}{2}\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}A\ge\dfrac{27}{16}+15\left(x=-\dfrac{3}{2}\right)\\A\ge-\dfrac{95}{16}+15\left(x=\dfrac{1}{2}\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}A\ge\dfrac{267}{16}\left(x=-\dfrac{3}{2}\right)\\A\ge\dfrac{145}{16}\left(x=\dfrac{1}{2}\right)\end{matrix}\right.\)

\(\Rightarrow A\ge\dfrac{145}{16}\left(x=\dfrac{1}{2}\right)\)

\(\Rightarrow GTNN\left(A\right)=\dfrac{145}{16}\left(x=\dfrac{1}{2}\right)\)