3x+x3=121
a) (x3 – x2 + x)(121 – 25y2 – 10y) – (x3 – x2 + x) – (121 – 25y2 – 10y) +1
X4 – 14x3 + 71x2 – 154x + 120
Giúp mik với ạ!!!!
a) (x3 – x2 + x)(121 – 25y2 – 10y) – (x3 – x2 + x) – (121 – 25y2 – 10y) +1
b) X4 – 14x3 + 71x2 – 154x + 120
Giúp mik vs
a: \(\left(x^3-x^2+x\right)\left(121-25y^2-10y\right)-\left(x^3-x^2+x\right)-\left(121-25y^2-10y\right)+1\)
\(=\left(x^3-x^2+x\right)\left(120-25y^2-10y\right)-\left(120-25y^2-10y\right)\)
\(=\left(120-25y^2-10y\right)\left(x^3-x^2+x-1\right)\)
\(=-\left[\left(25y^2+10y+1\right)-121\right]\left[x^2\left(x-1\right)+\left(x-1\right)\right]\)
\(=-\left(5y-10\right)\left(5y-12\right)\left(x-1\right)\left(x^2+1\right)\)
\(=-5\left(y-2\right)\left(5y-12\right)\left(x-1\right)\left(x^2+1\right)\)
b: \(x^4-14x^3+71x^2-154x+120\)
\(=x^4-5x^3-9x^3+45x^2+26x^2-130x-24x+120\)
\(=\left(x-5\right)\left(x^3-9x^2+26x-24\right)\)
\(=\left(x-5\right)\left(x^3-4x^2-5x^2+20x+6x-24\right)\)
\(=\left(x-5\right)\left(x-4\right)\left(x^2-5x+6\right)\)
\(=\left(x-5\right)\left(x-4\right)\left(x-3\right)\left(x-2\right)\)
[14/9: 2/121]x3/77
`@` `\text {Ans}`
`\downarrow`
\(\left[\dfrac{14}{9}\div\dfrac{2}{121}\right]\times\dfrac{3}{77}\)
`=`\(\dfrac{847}{9}\times\dfrac{3}{77}\)
`=`\(\dfrac{11}{3}\times\dfrac{1}{1}\)
`=`\(\dfrac{11}{3}\)
22 /5 x 6 / 121 x 11/4 x3/5 x 1/3 x5 /4
\(\dfrac{22}{5}\times\dfrac{6}{121}\times\dfrac{11}{4}\times\dfrac{3}{5}\times\dfrac{1}{3}\times\dfrac{5}{4}\)
\(=\left(\dfrac{22}{5}\times\dfrac{5}{4}\right)\times\left(\dfrac{6}{121}\times\dfrac{11}{4}\right)\times\left(\dfrac{3}{5}\times\dfrac{1}{3}\right)\)
\(=\dfrac{11}{2}\times\dfrac{3}{22}\times\dfrac{1}{5}\)
\(=\dfrac{3}{20}\)
\(=\dfrac{22\times6\times11\times3\times1\times5}{5\times121\times4\times5\times3\times4}=\dfrac{11\times2\times6\times11\times1}{11\times11\times4\times5\times4}=\dfrac{2\times6\times1}{4\times5\times4}=\dfrac{18}{100}=\dfrac{9}{50}\)
Tính.
a, (x3-2x2-10x-7):(x2-7-3x)
b, (x3+4x2+8x+5):(x+1)
c, (x3-x2-13x-14):(x2-3x-7)
d, (x3+5x2+5x):(x+5)
a: \(=\dfrac{x^3-3x^2-7x+x^2-3x-7}{x^2-3x-7}=x+1\)
b:\(=\dfrac{x^3+x^2+3x^2+3x+5x+5}{x+1}=x^2+3x+5\)
c:\(=\dfrac{x^3-3x^2-7x+2x^2-6x-14}{x^2-3x-7}=x+2\)
d: \(=\dfrac{x^2\left(x+5\right)+5x+25-25}{x+5}=x^2+5-\dfrac{25}{x+5}\)
Cho x,yÎR thỏa mãn: 3 x 2 + y 2 - 2 . l o g 2 ( x - y ) = 1 2 1 + log 2 1 - x y . Tìm giá trị lớn nhất của M = 2 ( x 3 + y 3 ) - 3 x y .
A. 13 2
B. 17 2
C. 3
D. 7
a, abc4 - 4abc = 1872
b, 3ab x3 + 210 = 12ab
c, abc= 15abc : 121
d, abc x 6 = a0bc
Ta có : abc4 - 4abc = 1872
=> abc x 10 + 4 - 4000 - abc = 1872
=> abc x 9 - 3996 = 1872
=> abc x 9 = 5868
=> abc = 652
cảm ơn anh nhìu nha vừa mới đăng xong đã có câu trả lời rùi . THANK YOU , VERY MUCH
các bạn giúp mình giải câu c nhé mai mình nộp bài rồi
Bài 5: Giải các phương trình sau:
a. (3x - 1)2 - (x + 3)2 = 0
b. x3 = \(\dfrac{x}{49}\)
c. x2 - 7x + 12 = 0
d. 4x2 - 3x -1 = 0
e. x3 - 2x - 4 = 0
f. x3 + 8x2 + 17x +10 = 0
g. x3 + 3x2 + 6x + 4 = 0
h. x3 - 11x2 + 30x = 0
a. (3x - 1)2 - (x + 3)2 = 0
\(\Leftrightarrow\left(3x-1+x+3\right)\left(3x-1-x-3\right)=0\)
\(\Leftrightarrow\left(4x+2\right)\left(2x-4\right)=0\)
\(\Leftrightarrow4x+2=0\) hoặc \(2x-4=0\)
1. \(4x+2=0\Leftrightarrow4x=-2\Leftrightarrow x=-\dfrac{1}{2}\)
2. \(2x-4=0\Leftrightarrow2x=4\Leftrightarrow x=2\)
S=\(\left\{-\dfrac{1}{2};2\right\}\)
b. \(x^3=\dfrac{x}{49}\)
\(\Leftrightarrow49x^3=x\)
\(\Leftrightarrow49x^3-x=0\)
\(\Leftrightarrow x\left(49x^2-1\right)=0\)
\(\Leftrightarrow x\left(7x+1\right)\left(7x-1\right)=0\)
\(\Leftrightarrow x=0\) hoặc \(7x+1=0\) hoặc \(7x-1=0\)
1. x=0
2. \(7x+1=0\Leftrightarrow7x=-1\Leftrightarrow x=-\dfrac{1}{7}\)
3. \(7x-1=0\Leftrightarrow7x=1\Leftrightarrow x=\dfrac{1}{7}\)
*Cách khác:
a) Ta có: \(\left(3x-1\right)^2-\left(x+3\right)^2=0\)
\(\Leftrightarrow\left(3x-1\right)^2=\left(x+3\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=-x-3\\3x-1=x+3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=-2\\2x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=2\end{matrix}\right.\)
Vậy: \(S=\left\{-\dfrac{1}{2};2\right\}\)
Phân tích các đa thức sau thành nhân tử:
a,x3+4x-5
b,x3-3x2+4
c,x3+2x2+3x+2
d,x2+2xy+y2+2x-2y-3
e,(x2+3x)2-2(x2+3x)-8
f,(x2+4x+10)2-7(x2+4x+11)+7
a) x3+4x-5 = x3-x2+x2+4x-5=(x3-x2)+(x2-x)+(5x-5)=x2(x-1)+x(x-1)+5(x-1)=(x2+x+5)(x-1)
b) x3-3x2+4=x3-2x2-x2+4=(x3-2x2)-(x2-4)=x2(x-2)-(x-2)(x+2)=(x2-x+2)(x-2)
c) x3+2x2+3x+2=x3+x2+x2+x+2x+2=(x3+x2)+(x2+x)+(2x+2)=x2(x+1)+x(x+1)+2(x+1)=(x2+x+2)(x+1)
d) bạn xem lại đề đúng ko
e) (x2+3x)2-2(x2+3x)-8=x4+6x3+9x2-2x2-6x-8=x4+6x3+7x2-6x-8=x4-x3+7x3-7x2+14x2-14x+8x-8=(x4-x3)+(7x3-7x2)+(14x2-14x)+(8x-8)=x3(x-1)+7x2(x-1)+14x(x-1)+8(x-1)=(x3+7x2+14x+8)(x-1)=(x3+x2+6x2+6x+8x+8)(x-1)=\(\left[\left(x^3+x^2\right)+\left(6x^2+6x\right)+\left(8x+8\right)\right]\left(x-1\right)\)\(=\left[x^2\left(x+1\right)+6x\left(x+1\right)+8\left(x+1\right)\right]\left(x-1\right)\)\(=\left(x^2+6x+8\right)\left(x+1\right)\left(x-1\right)\)\(=\left(x^2+2x+4x+8\right)\left(x+1\right)\left(x-1\right)\)\(=\left[\left(x^2+2x\right)+\left(4x+8\right)\right]\left(x+1\right)\left(x-1\right)\)\(=\left[x\left(x+2\right)+4\left(x+2\right)\right]\left(x+1\right)\left(x-1\right)\)=\(\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x+4\right)\)
f) (x2+4x+10)2-7(x2+4x+11)+7=(x2+4x+10)2-\(\left[7\left(x^2+4x+11\right)-7\right]\)\(=\left(x^2+4x+10\right)^2-7\left(x^2+4x+10\right)\)\(=\left(x^2+4x+10\right)\left(x^2+4x+3\right)\)
a) Ta có: \(x^3+4x-5\)
\(=x^3-x+5x-5\)
\(=x\left(x-1\right)\left(x+1\right)+5\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+x+5\right)\)
b) Ta có: \(x^3-3x^2+4\)
\(=x^3+x^2-4x^2+4\)
\(=x^2\left(x+1\right)-4\left(x-1\right)\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-4x+4\right)\)
\(=\left(x+1\right)\cdot\left(x-2\right)^2\)
c) Ta có: \(x^3+2x^2+3x+2\)
\(=x^3+x^2+x^2+x+2x+2\)
\(=x^2\left(x+1\right)+x\left(x+1\right)+2\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+x+2\right)\)
d) Ta có: \(x^2+2xy+y^2+2x+2y-3\)
\(=\left(x+y\right)^2+2\left(x+y\right)-3\)
\(=\left(x+y\right)^2+3\left(x+y\right)-\left(x+y\right)-3\)
\(=\left(x+y\right)\left(x+y+3\right)-\left(x+y+3\right)\)
\(=\left(x+y+3\right)\left(x+y-1\right)\)
e) Ta có: \(\left(x^2+3x\right)^2-2\left(x^2+3x\right)-8\)
\(=\left(x^2+3x\right)^2-4\left(x^2+3x\right)+2\left(x^2+3x\right)-8\)
\(=\left(x^2+3x\right)\left(x^2+3x-4\right)+2\left(x^2+3x-4\right)\)
\(=\left(x^2+3x-4\right)\left(x^2+3x+2\right)\)
\(=\left(x+4\right)\left(x-1\right)\left(x+1\right)\left(x+2\right)\)
f) Ta có: \(\left(x^2+4x+10\right)^2-7\left(x^2+4x+11\right)+7\)
\(=\left(x^2+4x+10\right)^2-7\left(x^2+4x+10\right)-7+7\)
\(=\left(x^2+4x+10\right)\left(x^2+4x+10-7\right)\)
\(=\left(x^2+4x+3\right)\left(x^2+4x+10\right)\)
\(=\left(x+1\right)\left(x+3\right)\left(x^2+4x+10\right)\)