\(a)P\left(x\right)=5x^5+3x-4x^4-2x^3+6+4x^2\)

   \(P\left(x\right)=5x^5-4x^4-2x^3+4x^2+3x+6\)

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

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

\(a)P\left(x\right)-Q\left(x\right)=\left(5x^5-4x^4-2x^3+4x^2+3x+6\right)+\left(-x^5+2x^4-2x^3+3x^2-x+\dfrac{1}{4}\right)\)

\(P\left(x\right)-Q\left(x\right)=5x^5-4x^4-2x^3+4x^2+3x+6+x^5-2x^4+2x^3-3x^2+x-\dfrac{1}{4}\)

\(P\left(x\right)-Q\left(x\right)=\left(5x^5+x^5\right)+\left(-4x^4-2x^4\right)+\left(-2x^3+2x^3\right)+\left(4x^2-3x^2\right)+\left(3x+x\right)+\left(6-\dfrac{1}{4}\right)\)

\(P\left(x\right)-Q\left(x\right)=6x^5-6x^4+x^2+4x+\dfrac{23}{4}\)

\(\text{c)Thay x=-1 vào biểu thức P(x),ta được:}\)

\(P\left(x\right)=5.\left(-1\right)^5-4.\left(-1\right)^4-2.\left(-1\right)^3+4.\left(-1\right)^2+3.\left(-1\right)+6\)

\(P\left(x\right)=\left(-5\right)-4-\left(-2\right)+4+\left(-3\right)+6\)

\(P\left(x\right)=\left(-9\right)-\left(-2\right)+4+\left(-3\right)+6\)

\(P\left(x\right)=\left(-7\right)+4+\left(-3\right)+6\)

\(P\left(x\right)=\left(-3\right)+\left(-3\right)+6\)

\(P\left(x\right)=\left(-6\right)+6=0\)

\(\text{Vậy giá trị của P(x) tại x=-1 là:0}\)

\(\text{Vậy =-1 là nghiệm của P(x)}\)

\(\text{Thay x=-1 vào biểu thức Q(x),ta được:}\)

\(Q\left(x\right)=\left(-1\right).5+2.\left(-1\right)^4-2.\left(-1\right)^3+3.\left(-1\right)^2-\left(-1\right)+\dfrac{1}{4}\)

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

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

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

\(Q\left(x\right)=\left(-2\right)-\left(-1\right)+\dfrac{1}{4}\)

\(Q\left(x\right)=\left(-3\right)+\dfrac{1}{4}=\dfrac{-13}{4}\)

\(\text{Vậy x=-1 không phải là nghiệm của Q(x)}\)

\(\text{d)Thay x=-1 vào biểu thức }P\left(x\right)-Q\left(x\right),\text{ta được:}\)

\(P\left(x\right)-Q\left(x\right)=6.\left(-1\right)^5-6.\left(-1\right)^4+\left(-1\right)^2+4.\left(-1\right)+\dfrac{23}{4}\)

\(P\left(x\right)-Q\left(x\right)=\left(-6\right)-6+1+\left(-4\right)+\dfrac{23}{4}\)

\(P\left(x\right)-Q\left(x\right)=\left(-12\right)+1+\left(-4\right)+\dfrac{23}{4}\)

\(P\left(x\right)-Q\left(x\right)=\left(-11\right)+\left(-4\right)+\dfrac{23}{4}\)

\(P\left(x\right)-Q\left(x\right)=\left(-15\right)+\dfrac{23}{4}=\dfrac{-37}{4}\)

\(\text{Vậy giá trị của P(x)-Q(x) tại x=-1 là:}\dfrac{-37}{4}\)

\(\dfrac{-7}{x+1}=\dfrac{6}{x+27}\)

\(\Rightarrow-7x-189=6x+6\)

\(\Rightarrow-7x-6x=6+189\)

\(\Rightarrow-13x=195\)

\(\Rightarrow x=-15\)

Vay x = -15

\(\dfrac{-7}{x+1}=\dfrac{6}{x+27}\)

\(\Rightarrow-7\left(x+27\right)=6\left(x+1\right)\)

\(\Rightarrow-7x-189=6x+6\)

\(\Rightarrow-7x-6x=6+189\)

\(\Rightarrow-13x=195\)

\(\Rightarrow x=-15\)

Vậy x=-15

\(4,7\div0,25+5,3\times4\)

\(=18,8+21,2\)

\(=40\)

\(3\times\left(a-2\right)+150=240\)

\(3\times\left(a-2\right)=90\)

\(a-2=30\)

\(a=32\)

\(\dfrac{1}{9}+a+\dfrac{7}{12}=\dfrac{17}{18}\)

\(\dfrac{1}{9}+a=\dfrac{13}{36}\)

\(a=\dfrac{1}{4}\)

\(\left(\dfrac{1}{2}\times\dfrac{1}{3}+\dfrac{1}{3}\times\dfrac{1}{4}+\dfrac{1}{4}\times\dfrac{1}{5}+\dfrac{1}{5}\times\dfrac{1}{6}+\dfrac{1}{6}\times\dfrac{1}{7}+\dfrac{1}{7}\times\dfrac{1}{8}\right)\times a=\dfrac{9}{16}\)

\(\left(\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+\dfrac{1}{6\times7}+\dfrac{1}{7\times8}\right)\times a=\dfrac{9}{16}\)

\(\left(\dfrac{1}{2}-\dfrac{1}{8}\right)\times a=\dfrac{9}{16}\)

\(\dfrac{3}{8}\times a=\dfrac{9}{16}\)

\(a=\dfrac{3}{2}\)

toán 5 mà khó thế bn

\(A=\dfrac{1}{2}\times\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{2011}-\dfrac{1}{2013}\right)=\dfrac{1}{2}\times\left(\dfrac{1}{5}-\dfrac{1}{2013}\right)=\dfrac{1004}{10065}\)

`(2x^2 +5)(3x-2)=0`

`<=> 2x^2 +5=0` hoặc `3x-2=0`

`<=> 2x^2 =-5` (vô lí vì `2x^2 ≥0 ∀x`) hoặc `x=2/3`

vậy pt có tập nghiệm `S={2/3}`

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

\(\Leftrightarrow\left[{}\begin{matrix}2x^2=-5\\3x=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2=-10\left(l\right)\\x=\dfrac{2}{3}\left(n\right)\end{matrix}\right.\)

Vậy \(S=\left\{\dfrac{2}{3}\right\}\)

Áp dụng tính chất dãy tỉ số bằng nhau:\(\dfrac{\left(5z-3y\right)+\left(3x-2z\right)+\left(2y-5x\right)}{2+5+3}\)

=\(\dfrac{\left(3x-5x\right)+\left(-3y+2y\right)+\left(5z-2z\right)}{2+5+3}\)

=\(\dfrac{-2x-y+3z}{2+5+3}\)(???!!!!)

=\(\dfrac{-2x}{2}=\dfrac{-y}{5}=\dfrac{3z}{3}\)

=\(\dfrac{2}{-2x}=\dfrac{5}{-y}=\dfrac{3}{3z}\)

tớ xin chịu trận vì ko chứng minh được :(((

nó lại ra như thế này

\(\dfrac{19\times2\times5}{19\times3\times5}=\dfrac{2}{3}\)

đúng òi

Số số hạng là (476-1):5+1=96(số)

Tổng là (476+1)*96/2=22896

x:6=1234+98        

x:6=2214

x  =2214*6

x  =13284

\(1)\)\(x\div6=1234+980\)

        \(x\div6=2214\)

        \(x\)     \(=2214\times6\)

       \(x\)     \(=13284\)

X:6=2214                                            X x (6+1) = 4585

X   =2214:6                                         X x 7        =4585

X   =369                                              X              =4585:7