a. (2x - 3)3 - (3x - 2).(3x + 2)
b. (5a -3).(5a +3) - (2a - 1)3
c. (x +2)3 - 6.(2x - 3)2 +12.(x + 2) - 8
d. (3x - 2017)3 + 3.(3x - 2017)2 + 3.(3x - 2017) +1
HELP me
Những câu hỏi liên quan
a. (2x - 3)3 - (3x -2).(3x+2)
b. (5x - 2)3 - (x2 -1).(x2 +1)
c. (5a - 3).(5a +3) - (2a - 1)3
d. (x + 2)3 - 6.(x +2)2 + 12.(x + 2) -8
e. (2x - 3)2 - 6.(2x - 3) +9
f. (3x -2017)3 +3. (3x - 2017)2 + 3.(3x - 2017) +1
Giải giúp mk nha k cho nè.
a. (2x - 3)3 - (3x - 2).(3x + 2)b. (4x + 3)3 - (5x + 1).(1 -5x)c. (5x - 2)3 - (x2 - 1) . (x2 +1)d. (3a - 2)3 - (4a + 1).(1 - 4a)e. (5a - 3).(5a +3) - (2a -1)3f. (9a - 2).(9a + 2) - (3a +1)3g. (3x + 5)3 - (8x -3).(8x +3)h.(x +2)3 - 6 (x +2)2 + 12.(x + 2) - 8i. (2x - 3)2 - 6 (2x - 3) + 9k. (3x - 2017)3 + 3.(3x - 2017)2 + 3.(3x - 2017) + 1giải giúp mk nha thks cần gấp sắp đi họk rồj HELP ME :(
Đọc tiếp
a. (2x - 3)3 - (3x - 2).(3x + 2)
b. (4x + 3)3 - (5x + 1).(1 -5x)
c. (5x - 2)3 - (x2 - 1) . (x2 +1)
d. (3a - 2)3 - (4a + 1).(1 - 4a)
e. (5a - 3).(5a +3) - (2a -1)3
f. (9a - 2).(9a + 2) - (3a +1)3
g. (3x + 5)3 - (8x -3).(8x +3)
h.(x +2)3 - 6 (x +2)2 + 12.(x + 2) - 8
i. (2x - 3)2 - 6 (2x - 3) + 9
k. (3x - 2017)3 + 3.(3x - 2017)2 + 3.(3x - 2017) + 1
giải giúp mk nha thks cần gấp sắp đi họk rồj HELP ME :(
tính giá trị của biểu thức:
\(P=\left(2x^5+2x^4-x^3-1\right)^{2016}+\left(\sqrt{2x+2x-3x+3x+3}\right)^3+\dfrac{\left(2x^3+2x^2-x-3\right)^{2017}}{2x^4+2x^3-x^2-3^{2017}}\)
khi \(x=\sqrt{\dfrac{2-\sqrt{3}}{2}}\)
giải pt (3x/x^2-x+3)-(2x/x^2-3x+3)=1
help me m cần gấp ạ
Đặt x/(x^2-3x+3) = t ta được
\(3t-2t=1\Leftrightarrow t=1\)
Theo cách đặt \(x=x^2-3x+3\Leftrightarrow x^2-4x+3=0\)
\(\Leftrightarrow\left(x-2\right)^2-1=0\Leftrightarrow\left(x-3\right)\left(x-1\right)=0\Leftrightarrow x=3;x=1\)
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Tìm GTNN
2) B= /3x-1/+/4y+2/-3x
3) C= /2x-1/+/2x+5/+2017
4) D= /x+3/+/2x-1/+/x-1/
Tập nghiệm của bất pt
a) \(\sqrt{x-2017}>\sqrt{2017-x}\)
b) \(\dfrac{2x^2-3x+4}{x^2+3}>2\)
c) \(3-2x+\sqrt{2-x}< x+\sqrt{2-x}\)
a, ĐK: \(x=2017\)
\(\sqrt{x-2017}>\sqrt{2017-x}\)
\(\Leftrightarrow\left\{{}\begin{matrix}2017-x\ge0\\x-2017>2017-x\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le2017\\x>2017\end{matrix}\right.\)
\(\Rightarrow S=\varnothing\)
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b, \(\dfrac{2x^2-3x+4}{x^2+3}>2\)
\(\Leftrightarrow2x^2-3x+4>2x^2+6\)
\(\Leftrightarrow x< -\dfrac{2}{3}\)
\(\Rightarrow S=\left(-\infty;-\dfrac{2}{3}\right)\)
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c, ĐK: \(x\le2\)
\(3-2x+\sqrt{2-x}< x+\sqrt{2-x}\)
\(\Leftrightarrow3-2x+\sqrt{2-x}< x+\sqrt{2-x}\)
\(\Leftrightarrow3x>3\)
\(\Leftrightarrow x>1\)
\(\Rightarrow S=(1;2]\)
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tim x , biet :
1 ) 2017 - \ x - 2017 \ = x
2) -1/2 . ( 3x - 1 ) + 3/4 . ( 3-2x) = -3 . ( x/2 - 1 ) - ( 4/5) mu -1
Cho \(x^2-x-1=0\).Tính \(Q=\frac{x^6-3x^5+3x^4-x^3+2017}{x^6-x^3-3x^2-3x+2017}\)
Ta có \(x^2-x-1=0\Rightarrow x^2-x=1\Rightarrow\left(x^2-x\right)^3=1\)
\(\Rightarrow x^6-3x^5+3x^4-x^3=1\)
Mặt khác \(x^2-x-1-0\Rightarrow x^2=x+1\)
\(\Rightarrow x^6=\left(x+1\right)^3=x^3+2=3x^2+3x+1\)
\(\Rightarrow P=\frac{1+2017}{1+2017}=1\)
tìm x:
a)3(2x-3)+2(2-x)=-3
b)2x(x2-2)+x2(1-2x)-x2=-12
c)3x(2x+3)-(2x+5)(3x-2)=8
d)4x(x - 1) - 3(x2-5)-x2=(x-3)-(x+4)
e)2(3x-1)(2x+5)-6(2x-1)(x+2)=-6
a: Ta có: \(3\left(2x-3\right)+2\left(2-x\right)=-3\)
\(\Leftrightarrow6x-9+4-2x=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
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