câu a sai đề nha

Nếu câu a đề đúng thì phương trình vô nghiệm nha

Theo mình đây là đề đúng

\(\left(2x^2+3x-1\right)^2-5\left(2x^2+3x-1\right)-24=0\)

Đặt a=\(\left(2x^2+3x-1\right)\)

Khi đó, phương trình trở thành

\(a^2-5a-24=0\)

\(\left(a-8\right)\left(a+3\right)=0\)

\(\left[{}\begin{matrix}a=8\\a=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x^2+3x-1=8\\2x^2+3x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x^2+3x-9=0\\2x^2+3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left(x+3\right)\left(2x-3\right)=0\\2\left(x+\dfrac{3}{2}\right)^2+\dfrac{7}{8}=0\left(vl\right)\end{matrix}\right.\)\(\left[{}\begin{matrix}x=-3\\x=\dfrac{3}{2}\end{matrix}\right.\)

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

\(\Leftrightarrow6x^3-9x^2-9x+6=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+1\right)\left(x-\dfrac{1}{2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\\x=\dfrac{1}{2}\end{matrix}\right.\)

1.\(x^{16}-y^{16}=\left(x^8-y^8\right)\left(x^8+y^8\right)\)

2.\(x^3-125=x^3-5^3=\left(x-5\right)\left(x^2+5x+25\right)\)

\(-64+\frac{1}{8}x^3=\left(\frac{x}{2}\right)^3-4^3=\left(\frac{x}{2}-4\right)\left(\frac{x^2}{4}+2x+16\right)\)

\(8x^3+60x^2y+150xy^2+125y^3=\left(2x\right)^3+3.\left(2x\right)^2.\left(5y\right)+3.\left(2x\right).\left(5y\right)^2+\left(5y\right)^3\)

\(=\left(2x+5y\right)^3\)

cám ơn bn Nguyễn Minh Quang nhé

Cân lun!

\(x^4-10x^3-15x^2+20x+4\)

\(=x^4-x^3-9x^3+9x^2-24x^2+24x-4x+4\)

\(=x^3\left(x-1\right)-9x^2\left(x-1\right)-24x\left(x-1\right)-4\left(x-1\right)\)

\(=\left(x-1\right)\left(x^3-9x^2-24x-4\right)\)

\(=\left(x-1\right)\left(x^3+2x^2-11x^2-22x-2x-4\right)\)

\(=\left(x-1\right)\left[x^2\left(x+2\right)-11x\left(x+2\right)-2\left(x+2\right)\right]\)

\(=\left(x-1\right)\left(x+2\right)\left(x^2-11x-2\right)\)

Chúc bạn học tốt!!!

a) Đặt \(x^2=y\Rightarrow x^4+x^2-20=y^2+y-20=y^2-4y+5y-20=\left(y-4\right)\left(y+5\right)\)

Thay trở lại, ta có: \(x^4+x^2-20=\left(x^2-4\right)\left(x^2+5\right)=\left(x-2\right)\left(x+2\right)\left(x^2+5\right)\)

b) Đặt \(x-y=z\Rightarrow\left(x-y\right)^2+4x-4y-12=z^2+4z-12=z^2-2z+6z-12=\left(z-2\right)\left(z+6\right)\)

Thay trở lại ta có kết quả sau: \(\left(x-y-2\right)\left(x-y+6\right)\)

a) \(\frac{1+\frac{1}{x}}{x-\frac{1}{x}}=\frac{x+1}{x}\div\frac{x^2-1}{x}=\frac{x+1}{x}\cdot\frac{x}{\left(x+1\right)\left(x-1\right)}=\frac{1}{x-1}\)

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

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

a) \(\frac{1+\frac{1}{x}}{x-\frac{1}{x}}=\frac{\frac{x+1}{x}}{\frac{x^2-1}{x}}=\frac{x+1}{x}\cdot\frac{x}{x^2-1}=\frac{1}{x-1}\)

b) \(\left(\frac{1}{\left(x+2\right)^2}-\frac{1}{\left(x-2^2\right)}\right):\left(\frac{1}{x+2}+\frac{1}{x-2}\right)\)

\(\Leftrightarrow\left(\frac{\left(x-2\right)^2-\left(x+2\right)^2}{\left(x+2\right)^2\left(x-2\right)^2}\right):\left(\frac{1}{x+2}+\frac{1}{x-2}\right)\)

\(\Leftrightarrow\left(\frac{x^2-4x+4-x^2-4x-4}{\left[\left(x-2\right)\left(x+2\right)\right]^2}\right):\left(\frac{x-2+x+2}{x^2-4}\right)\)

\(\Leftrightarrow\frac{-8x}{\left(x^2-4\right)^2}\cdot\frac{x^2-4}{2x}\)\(\Leftrightarrow-\frac{4}{x^2-4}\)

d) \(\frac{3x}{x^3-1}+\frac{x-1}{x^2+x+1}\Leftrightarrow\frac{3x}{x^3-1}+\frac{\left(x-1\right)^2}{x^3-1}\)

\(\Leftrightarrow\frac{x^2-2x+1+3x}{x^3-1}=\frac{x^2+x+1}{x^3-1}=\frac{1}{x-1}\)

còn lại chút giải tiếp !!!

1/ (x² - 2)(x² + 2x + 2)

2/ x² - (x² + 2)² = (x - x² - 2)(x + x² ​+ 2)

Bài 1:

a. A = x^2 - 5x - 1

\(=x^2-5x+\frac{25}{4}-\frac{29}{4}\)

\(=x^2-5x+\left(\frac{5}{2}\right)^2-\frac{29}{4}\)

\(=\left(x-\frac{5}{2}\right)^2-\frac{29}{4}\ge0-\frac{29}{4}=-\frac{29}{4}\)

Dấu = khi x=5/2

Vậy MinC=-29/4 khi x=5/2

2. Tìm x:
a. ( 2x - 3 )^2 - ( 4x + 1 )( 4x - 1 ) = ( 2x - 1 ).( 3 - 7x )

=>4x²-12x+9+1-16x²=-14x²+13x-3

=>-12x²-12x+10=-14x²+13x-3

=>2x²-25x+13=0

\(\Rightarrow2\left(x-\frac{25}{4}\right)^2-\frac{521}{8}=0\)

\(\Rightarrow\left(x-\frac{25}{4}\right)^2=\frac{521}{16}\)

\(\Rightarrow x-\frac{25}{4}=\pm\sqrt{\frac{521}{16}}\)

\(\Rightarrow x=\frac{25}{4}\pm\frac{\sqrt{521}}{4}\)

c. 4.( x - 3 ) - ( x + 2 ) = 0

=>4x-12-x-2=0

=>3x-14=0

=>3x=14

=>x=14/3

Bạn làm bài kiểm tra hả sao nhiều bài tek. Mk làm mất khá nhiều tg luôn đó

Có một số câu thì mình không làm được. Mong bạn thông cảm!!!

a) y(x²-y²)(x²+y²)-y(x⁴-y⁴)=y[(x²)²-(y²)²] - y(x⁴-y⁴)=y(x⁴-y⁴)-y(x⁴-y⁴)=0

vậy giá trị biểu thức không phụ thuộc vào biến (đpcm)

b) \(\left(\frac{1}{3}+2x\right)\left(4x^2-\frac{2}{3}x+\frac{1}{9}\right)-\left(8x^3-\frac{1}{27}\right)\)

\(=\left[\left(2x\right)^3+\left(\frac{1}{3}\right)^3\right]-\left(8x^3-\frac{1}{27}\right)=8x^3+\frac{1}{27}-8x^3+\frac{1}{27}=\frac{1}{54}\)

vậy giá trị biểu thức không phụ thuộc vào biến (đpcm)

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

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

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

= 0 (đpcm)