x² + 2x = 0

=> x(x + 2) = 0

=> \(\orbr{\begin{cases}x=0\\x+2=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=0\\x=-2\end{cases}}\)

(x - 2) + 3.x² - 6x = 0

=> (x - 2) + 3x² - 3x . 2 = 0

=> (x - 2) + 3x.(x - 2) = 0

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

=> \(\orbr{\begin{cases}1+3x=0\\x-2=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=-\frac{1}{3}\\x=2\end{cases}}\)

M+N=(3/2x6-7x+4x^5+2,5x^2)+(-3x^6+1/2^5-13/2x^2+4x)

M+N=3/2x6-7x+4x^5+2,5x^2+-3x^6+1/2^5-13/2x^2+4x

= (3/2x^6-3x^6)+(7x+4x)+(4x^5+1/2^5)+(2,5x^2-13/2x^2)

=-1,5x^6+11x+4,5x^5-4x^2

M-N=(3/2^6-7x+4x^5+2,5x^2)-(-3x^6+1/2^5-13/2x^2+4x)

=3/2^6-7x+4x^5+2,5x^2+3x^6-1/2^5+13/2x^2-4x

= (3/2x^6+3x^6)+(-7x-4x)+(4x^5-1/2^5)+(2,5x^2+13/2x^2)

= 4,5x^6-11x+3,5x^5+9x^2

N-M=(-3x^6+1/2^5-13/2x^2+4x)-(3/2^6-7x+4x^5+2,5x^2)

= -3x^6+1/2^5-13/2x^2+4x-3/2^6-7x-4x^5-2,5x^2

= (-3x^6-3/2x^6)+(1/2x^5-4x^5)+(-13/2x^2-2,5x^2)+(4x-7x)

= -4,5x^6-3,5x^5-9x^2-3x

Cho đa thức

P(x)= x mũ 2 + 2x mũ 2 +1 (1)

Thay P(-1) vào đa thức (1) , ta có :

P= \((-1)^2 +2.(-1) ^3\)

P= \(1+ (-2)\)

P= \(-1\)

Thay P(\(\dfrac{1}{2}\)) vào đa thức (1) , ta có :

\(P= (\dfrac{1}{2})^2 +2.(\dfrac{1}{2})^3\)

\(P= \dfrac{1}{4} + \dfrac{1}{4}\)

\(P=\dfrac{1}{2}\)

Q(x)=x mũ 4 +4x mũ 3 +2x mũ 2 trừ 4x+ 1. (2)

Thay Q(-2) vào đa thức (2) , ta có :

Q =\((-2)^4 +4.(-2)^3 +2.(-2)^2-4(-2)+1\)

\(Q = 16-32+8+8+1\)

\(Q= 1\)

Thay Q(1) vào đa thức (2) , ta có:

\(Q= \) \(1^4+4.1^3+2.1^2-4.1+1\)

\(Q= 1+ 4+2-4+1\)

\(Q= 4\)

Tính P(-1) ; P(1/2) ; Q(-2) ; Q(1)

Bài 1:

a) Ta có: \(2x=5y.\)

=> \(\frac{x}{y}=\frac{5}{2}\)

=> \(\frac{x}{5}=\frac{y}{2}\) và \(x.y=90.\)

Đặt \(\frac{x}{5}=\frac{y}{2}=k\Rightarrow\left\{{}\begin{matrix}x=5k\\y=2k\end{matrix}\right.\)

Có: \(x.y=90\)

=> \(5k.2k=90\)

=> \(10k^2=90\)

=> \(k^2=90:10\)

=> \(k^2=9\)

=> \(k=\pm3.\)

TH1: \(k=3\)

\(\Rightarrow\left\{{}\begin{matrix}x=3.5=15\\y=3.2=6\end{matrix}\right.\)

TH2: \(k=-3\)

\(\Rightarrow\left\{{}\begin{matrix}x=\left(-3\right).5=-15\\y=\left(-3\right).2=-6\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(15;6\right),\left(-15;-6\right).\)

e) Ta có: \(\frac{x}{y}=\frac{4}{5}.\)

=> \(\frac{x}{4}=\frac{y}{5}\) và \(x.y=20.\)

Đặt \(\frac{x}{4}=\frac{y}{5}=k\Rightarrow\left\{{}\begin{matrix}x=4k\\y=5k\end{matrix}\right.\)

Có: \(x.y=20\)

=> \(4k.5k=20\)

=> \(20k^2=20\)

=> \(k^2=20:20\)

=> \(k^2=1\)

=> \(k=\pm1.\)

TH1: \(k=1\)

\(\Rightarrow\left\{{}\begin{matrix}x=1.4=4\\y=1.5=5\end{matrix}\right.\)

TH2: \(k=-1\)

\(\Rightarrow\left\{{}\begin{matrix}x=\left(-1\right).4=-4\\y=\left(-1\right).5=-5\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(4;5\right),\left(-4;-5\right).\)

Chúc bạn học tốt!

sao ngắn vậy bạn

ỏ cảm mơn nhaaaa ! có j giúp típ nha thank kiuuu 

\(2^3+\left(\dfrac{1}{5}\right)^4+5^4=8+\dfrac{1}{625}+625=\dfrac{5000+1+625^2}{625}=\dfrac{395626}{625}\)

\(Sửa:2^3+\left(\dfrac{1}{5}\right)^4\cdot5^4=8+\left(\dfrac{1}{5}\cdot5\right)^4=8+1=9\)

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

                                                                   Bài giải

a, \(\frac{2}{7}x+\frac{1}{2}=-\frac{3}{4}\)

\(\frac{2}{7}x=-\frac{3}{4}-\frac{1}{2}\)

\(\frac{2}{7}x=-\frac{5}{4}\)

\(x=-\frac{5}{4}\text{ : }\frac{2}{7}\)

\(x=-\frac{35}{8}\)

b, \(\left(6x+\frac{2}{5}\right)=-\frac{8}{125}\)

\(6x=-\frac{8}{125}-\frac{2}{5}\)

\(6x=-\frac{58}{125}\)

\(x=-\frac{58}{125}\text{ : }6\)

\(x=\frac{-29}{375}\)

c, \(\left|x-\frac{2}{3}\right|\cdot\left(18-6x^2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}\left|x-\frac{2}{3}\right|=0\\18-6x^2=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x-\frac{2}{3}=0\\6x^2=18\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x^2=3\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=\sqrt{3}\end{cases}}\)

\(\Rightarrow\text{ }x\in\left\{\frac{2}{3}\text{ ; }\sqrt{3}\right\}\)