2x+2x+2+2x+4=336
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Tìm x, y, z. Biết 2x = 3y - 3x = -7x và x.z = -336
$2x=-7x$ thì $x=0$ rồi bạn, làm sao mà $xz=-336$ hả bạn?
Rút gọn :
1. (2x-5)(3x+1)-(x-3)^2+(2x+5)^2-(3x+1)^3
2. (2x-1)(2x+1)-3x-2)(2x+3)-(x-1)^3+(2x+3)^3
3. (x-2)(x^2+2x+4)-(3x-2)^3+(3x-4)^2
4. (7x-1)(8x+2)-(2x-7)^2-(x-4)^3-(3x+1)^3
5. (5x-1)(5x+1)-(x+3)(x^2-3x+9)-(2x+4)^2-(3x-4)^2+(2x-5)^3
6. (4x-1)(x+2)-(2x+5)^2-(3x-7)^2+(2x+3)^3=(3x-1)^3
1: \(=6x^2+2x-15x-5-x^2+6x-9+4x^2+20x+25-27x^3-27x^2-9x-1\)
=-27x^3-18x^2+4x+10
2: =4x^2-1-6x^2-9x+4x+6-x^3+3x^2-3x+1+8x^3+36x^2+54x+27
=7x^3+37x^2+46x+33
5:
\(=25x^2-1-x^3-27-4x^2-16x-16-9x^2+24x-16+\left(2x-5\right)^3\)
\(=8x^3-60x^2+150-125+12x^2-x^3+8x-60\)
=7x^3-48x^2+8x-35
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cmr:1-2/x-(2x+x^2/4+2x+x^2 + 2x-x^2/4-2x+x^2):(16-8x/4-2x+x^2 -16+8x/4+2x+x^2)=(x-1/x)^2
(2x-1/x)^4
(2x^2-1/x)^4
[(2x)^2-1/x]^4
[(2x)^2-1/x^2]^4
(2x-1/x)^4
(2x^2-1/x)^4
[(2x)^2-1/x]^4
[(2x)^2-1/x^2]^4
(2x-1/x)^4
(2x^2-1/x)^4
[(2x)^2-1/x]^4
[(2x)^2-1/x^2]^4
(2x-1/x)^4
(2x^2-1/x)^4
[(2x)^2-1/x]^4
[(2x)^2-1/x^2]^4
làm phép chia :
a) (x^4 -2x^3 + 2x -1) : (x^2 - 1)
b) (x^3 -8) : (x^2 + 2x +4)
c) (x^6 - 2x^5 + 2x^4 + 6x^3 - 4x^2)n: 6x^2
d) (-2x^5 + 3x^2 - 4x^3) :2x^2
e) (15x^3 - 10x^2 + x - 2) : (x - 2)
f) (2x^4 - 3x^3 - 3x^2 + 6x - 2) : (x^2 - 2)
b: =x-2
d: \(=-x^3+\dfrac{3}{2}-2x\)
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H24
30 tháng 1 2021 lúc 20:26
4/2x+1-2/2x+3=1/2x+4-3/2x+2
Xem chi tiết
ĐKXĐ: \(x\notin\left\{-\dfrac{1}{2};-1;\dfrac{-3}{2};-2\right\}\)
Ta có: \(\dfrac{4}{2x+1}-\dfrac{2}{2x+3}=\dfrac{1}{2x+4}-\dfrac{3}{2x+2}\)
\(\Leftrightarrow\dfrac{4\left(2x+3\right)}{\left(2x+1\right)\left(2x+3\right)}-\dfrac{2\left(2x+1\right)}{\left(2x+1\right)\left(2x+3\right)}=\dfrac{2x+2}{\left(2x+2\right)\left(2x+4\right)}-\dfrac{3\left(2x+4\right)}{\left(2x+2\right)\left(2x+4\right)}\)
\(\Leftrightarrow\dfrac{8x+12-4x-2}{\left(2x+1\right)\left(2x+3\right)}=\dfrac{2x+2-6x-12}{\left(2x+2\right)\left(2x+4\right)}\)
\(\Leftrightarrow\dfrac{4x+10}{\left(2x+1\right)\left(2x+3\right)}=\dfrac{-4x-10}{\left(2x+2\right)\left(2x+4\right)}\)
\(\Leftrightarrow\dfrac{4x+10}{\left(2x+1\right)\left(2x+3\right)}-\dfrac{-4x-10}{\left(2x+2\right)\left(2x+4\right)}=0\)
\(\Leftrightarrow\dfrac{4x+10}{\left(2x+1\right)\left(2x+3\right)}+\dfrac{4x+10}{\left(2x+2\right)\left(2x+4\right)}=0\)
\(\Leftrightarrow\left(4x+10\right)\left(\dfrac{1}{\left(2x+1\right)\left(2x+3\right)}+\dfrac{1}{\left(2x+2\right)\left(2x+4\right)}\right)=0\)
\(\Leftrightarrow2\left(2x+5\right)\left(\dfrac{\left(2x+2\right)\left(2x+4\right)}{\left(2x+1\right)\left(2x+2\right)\left(2x+3\right)\left(2x+4\right)}+\dfrac{\left(2x+1\right)\left(2x+3\right)}{\left(2x+1\right)\left(2x+2\right)\left(2x+3\right)\left(2x+4\right)}\right)=0\)
\(\Leftrightarrow\left(2x+5\right)\left(4x^2+8x+4x+8+4x^2+6x+2x+6\right)=0\)(Vì \(\left(2x+1\right)\left(2x+2\right)\left(2x+3\right)\left(2x+4\right)\ne0\forall x\) thỏa mãn ĐKXĐ)
\(\Leftrightarrow\left(2x+5\right)\left(8x^2+20x+14\right)=0\)
mà \(8x^2+20x+14>0\forall x\)
nên 2x+5=0
\(\Leftrightarrow2x=-5\)
\(\Leftrightarrow x=-\dfrac{5}{2}\)
Vậy: \(S=\left\{-\dfrac{5}{2}\right\}\)
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\(\dfrac{4}{2x+1}-\dfrac{2}{2x+3}=\dfrac{1}{2x+4}-\dfrac{3}{2x+2}\)
\(\Leftrightarrow\dfrac{2x+5}{2x+1}-\dfrac{2x+5}{2x+3}=\dfrac{2x+5}{2x+4}-\dfrac{2x+5}{2x+2}\)
\(\Leftrightarrow\left(2x+5\right)\left(\dfrac{1}{2x+1}-\dfrac{1}{2x+3}-\dfrac{1}{2x+4}+\dfrac{1}{2x+2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{5}{2}\\\dfrac{1}{2x+1}-\dfrac{1}{2x+3}-\dfrac{1}{2x+4}+\dfrac{1}{2x+2}=0\left(1\right)\end{matrix}\right.\)
(1) \(\Leftrightarrow\dfrac{2x+3-2x-1}{\left(2x+1\right)\left(2x+3\right)}+\dfrac{2x+4-2x-2}{\left(2x+4\right)\left(2x+2\right)}=0\)
\(\Leftrightarrow\dfrac{2}{\left(2x+1\right)\left(2x+3\right)}+\dfrac{2}{\left(2x+4\right)\left(2x+2\right)}=0\)
\(\Leftrightarrow\dfrac{1}{\left(2x+1\right)\left(2x+3\right)}=-\dfrac{1}{\left(2x+4\right)\left(2x+2\right)}\)
......
(Vô lí)
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1. 2x*(3x^2+4)-6x*(x-7)
2. (x+1)*(2x-2)-(2x-1)^2
3. (2x-3)^2-2*(2x-3)*(2x+5)+(2x+5)^2
4. (x-2)^4+(x+2)^4
5. (x+1)^5-(x-1)^5
tách 2,3 câu ra làm 1 câu hỏi đi. bạn đăng cả đóng thế này k ai tl cho đâu. khi nào tách thì gửi link mình tl cho
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