bai 1

1 thay k=0 vao pt ta co 4x^2-25+0^2+4*0*x=0

<=>(2x)^2-5^2=0

<=>(2x+5)*(2x-5)=0

<=>2x+5=0 hoăc 2x-5 =0 tiếp tục giải ý 2 tương tự

Thêm nữa câu a) Tính: M(x) + N(x)+ P(x)

B) Tính M(x) - N (x) - P(x)

ok rồi giúp mình với nha

a/ \(\dfrac{3}{2}\left(x-\dfrac{5}{3}\right)+\dfrac{4}{5}=x+1\)

\(\Rightarrow\dfrac{3}{2}x-\dfrac{5}{2}-x=1+\dfrac{4}{5}\)

\(\Rightarrow\dfrac{3}{2}x-x=\dfrac{9}{5}+\dfrac{5}{2}\)

\(\Rightarrow\dfrac{1}{2}x=\dfrac{43}{10}\)

\(\Rightarrow x=\dfrac{43}{5}\)

b/ \(\dfrac{1}{6}\left(2x-3\right)=\dfrac{1}{2}\left(-x+\dfrac{1}{4}\right)-\dfrac{2}{3}\)

\(\Rightarrow\dfrac{1}{3}x-\dfrac{1}{2}=-\dfrac{1}{2}x+\dfrac{1}{8}-\dfrac{2}{3}\)

\(\Rightarrow\dfrac{1}{3}x+\dfrac{1}{2}x=\dfrac{1}{8}-\dfrac{2}{3}+\dfrac{1}{2}\)

\(\Rightarrow\dfrac{5}{6}x=-\dfrac{1}{24}\Rightarrow x=-\dfrac{1}{20}\)

c/ làm như b

d/ \(\left(x-1\right)^4=\left(x-1\right)^6\)

\(\Rightarrow\left[{}\begin{matrix}x-1=-1\\x-1=0\\x-1=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=2\end{matrix}\right.\)

Vậy \(x\in\left\{0;1;2\right\}\)

ta có : x^5+2x^4+3x^3+3x^2+2x+1=0

\(\Leftrightarrow\)x^5+x^4+x^4+x^3+2x^3+2x^2+x^2+x+x+1=0

\(\Leftrightarrow\)(x^5+x^4)+(x^4+x^3)+(2x^3+2x^2)+(x^2+x)+(x+1)=0

\(\Leftrightarrow\)x^4(x+1)+x^3(x+1)+2x^2(x+1)+x(x+1)+(x+1)=0

\(\Leftrightarrow\)(x+1)(x^4+x^3+2x^2+x+1)=0

\(\Leftrightarrow\)(x+1)(x^4+x^3+x^2+x^2+x+1)=0

\(\Leftrightarrow\)(x+1)[x^2(x^2+x+1)+(x^2+x+1)]=0

\(\Leftrightarrow\)(x+1)(x^2+x+1)(x^2+1)=0

VÌ x^2+x+1=(x+\(\dfrac{1}{2}\))^2+\(\dfrac{3}{4}\)\(\ne0\) và x^2+1\(\ne0\)

\(\Rightarrow\)x+1=0

\(\Rightarrow\)x=-1

CÒN CÂU B TỰ LÀM (02042006)

b: x^4+3x^3-2x^2+x-3=0

=>x^4-x^3+4x^3-4x^2+2x^2-2x+3x-3=0

=>(x-1)(x^3+4x^2+2x+3)=0

=>x-1=0

=>x=1

Ta có : |1 - 5x| - 1 = 3

=> |1 - 5x| = 4

\(\Leftrightarrow\orbr{\begin{cases}1-5x=4\\1-5x=-4\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}5x=1-4\\5x=1+4\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}5x=3\\5x=5\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{5}\\x=1\end{cases}}\)

giúp với trí ơi

giúp nốt trí ui

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

\(=\left(2x+y\right)\left(2x-y\right)-\left(2x+y\right)=\left(2x+y\right)\left(2x-y-1\right)\)

b) \(9x^2-25y^2-6x+10y=\left(3x-5y\right)\left(3x-5y\right)-2\left(3x-5y\right)\)

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

c) \(x^3-2x^2+2x-1=x^3-1-2x^2+2x\)

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

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

d) \(x^4+2x^3-4x-4=x^4-4+2x^3-4x\)

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