Giúp tôi giải toán và làm văn

Đặt \(a=\frac{1}{4453};b=\frac{1}{1997}\)ta có :

\(5\frac{6}{4453}\cdot\frac{1}{1997}-\frac{2}{1997}\cdot2\frac{3}{4453}\)

\(=\left(5+6a\right)\cdot b-2b\left(2+3a\right)\)

\(=5b+6ab-4b-6ab\)

\(=b=\frac{1}{1997}\)

Thế mà còn trả lời :(

anh oi em moi hoc lop 7

Ta có: \(x+y+z=0\)

\(\Rightarrow\) \(\hept{\begin{cases}x+y=-z\\y+z=-x\\x+z=-y\end{cases}}\)

\(A=x\left(x+y\right)\left(x+z\right)=x\left(-z\right)\left(-y\right)=xyz\)

\(B=y\left(y+z\right)\left(y+x\right)=y\left(-x\right)\left(-z\right)=xyz\)

\(B=z\left(z+x\right)\left(y+z\right)=z\left(-y\right)\left(-x\right)=xyz\)

\(\Rightarrow A=B=C\)

Tham khảo nhé~

cái cuối dấu cộng mới biết làm,,

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

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

\(=\left(x^2-1\right)\left[\left(x+4\right)^2-1\right]\)

\(=\left(x-1\right)\left(x+1\right)\left(x+4+1\right)\left(x+4-1\right)\)

\(=\left(x-1\right)\left(x+1\right)\left(x+5\right)\left(x-3\right)\)

=.= hok tốt!!

\(\frac{1}{2}x+x^2y+xy\cdot2y-4\)

\(=\frac{1}{2}x+x^3y^3\cdot2y-4\)

\(=\frac{1}{2}x^4+2y^4-4\)

Vậy đa thức đã được thu gọn là  \(\frac{1}{2}x^4+2y^4-4\)

Ta có :

\(Q+P\)

\(=3xyz+2.5xy^2-2+2xyz+1.5xyz-y^3\)

\(=6.5xyz+2.5xy^2-y^3-2\)

\(Q-P=3xyz+2.5xy^2-2-2xyz-1.5xyz+y^3\)

\(=2.5xy^2-0.5xyz+y^3-2\)

\(P-Q=2xyz+1.5xyz-y^3-3xyz-2.5xy^2+2\)

\(=0.5xyz-y^3-2.5xy^2+2\)

\(M+N=\) \(5x^3y+9xy^2-7,5xyz+y^3\)

\(M-N=\) \(x^3-xy^2+0,5xyz+y^3\)

Chúc Bạn Học Tốt

\(3xyz^2+\left(-\frac{4}{8}\right)xyz^5\cdot\frac{1}{3}xyz\)

\(=3xyz^2-\frac{1}{2}xyz\cdot\frac{1}{3}xyz\)

\(=3xyz-\frac{1}{6}x^2y^2z^2\)

\(xyz\left(3-\frac{1}{6}xyz\right)\)

b) \(3xyz^5\cdot\left(-\frac{1}{7}\right)xyz\cdot\frac{-1}{8}xyz^4\)

\(=\left[3\cdot\left(-\frac{1}{7}\right)\cdot\left(-\frac{1}{8}\right)\right]\left(x\cdot x\cdot x\right)\left(y\cdot y\cdot y\right)\left(z^5\cdot z\cdot z^4\right)\)

\(=\frac{3}{56}x^3y^3z^{10}\)

a, \(3xyz^2+\left(\frac{-4}{8}xyz^5\right)\cdot\frac{1}{3}xyz=3xyz^2+\left[\left(\frac{-4}{8}\right)\cdot\frac{1}{3}\right]xyz^5xyz\)\(=3xyz^2-\frac{1}{2}x^2y^2z^6\)

b, \(3xyz^5\cdot\left(\frac{-1}{7}xyz^2\right)\cdot\frac{-1}{8}xyz^4=\left[3\cdot\left(\frac{-1}{7}\right)\cdot\left(\frac{-1}{8}\right)\right]xyz^5xyz^2xyz^4=\frac{3}{56}x^3y^3z^{11}\)

a) (2x - 5)(3x + b) = ax^2 + x + c
<=> 6x^2 + 2bx -15x -5b = ax^2 + x + c
<=> -ax^2 + 2bx -5b -c = -6x^2 +16x
Đồng nhất hệ số ta có :
+) -a = -6 => a= 6
+) 2b = 16 => b= 8
+) -5b -c= 0 => c= -40

c ) (ax+b)( x^2 -x-1)= ax^3 - cx^2 - 1
<=> ax^3 -ax^2-ax +bx^2-bx-b= ax^3 - cx^2 - 1
<=> (c+b-a)x^2 -(a+b)x -b = -1
Đồng nhất hệ số ta được:
+) c+b-a =0
+) -a-b = 0
+) -b = -1 => b= 1
Thay b=1 ta được a = -1 và c= -2

<p>a) (2x - 5)(3x + b) = ax^2 + x + c<br>&lt;=&gt; 6x^2 + 2bx -15x -5b =&nbsp;ax^2 + x + c<br>&lt;=&gt; -ax^2 + 2bx -5b -c = -6x^2 +16x<br>Đồng nhất hệ số ta có :<br>+) -a = -6 =&gt; a= 6<br>+) 2b = 16 =&gt; b= 8<br>+) -5b -c= 0 =&gt; c= -40</p>

P/s : Easy mà bạn :

Ta có : 

\(P\left(x\right)=ax^2+bx+c\)

\(\Rightarrow\hept{\begin{cases}P\left(-1\right)=a.\left(-1\right)^2+b.\left(-1\right)+c\\P\left(3\right)=a.3^2+b.3+c\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}P\left(-1\right)=a-b+c\\P\left(3\right)=9a+3b+c\end{cases}}\)

\(\Rightarrow P\left(3\right)-P\left(-1\right)=9a+3b+c-\left(a-b+c\right)\)

\(\Rightarrow P\left(3\right)-P\left(-1\right)=8a+4b\)

\(\Rightarrow P\left(3\right)-P\left(-1\right)=4\left(2a+b\right)\)

\(\Rightarrow P\left(3\right)-P\left(-1\right)=4.0=0\)

\(\Rightarrow P\left(3\right)=P\left(-1\right)\)

\(\Rightarrow\)

\(P\left(3\right).P\left(-1\right)=P\left(3\right).P\left(3\right)=\left[P\left(3\right)\right]^2\ge0\)

\(\left(Đcpm\right)\)