\(x^5-3x^4-\dfrac{1}{2}x^3-x^2-5-15x^6-0.75x^5+2x^3-x+8\)
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Y
26 tháng 11 2023
\(\dfrac{3}{x-5}-\dfrac{x+1}{x\left(x-5\right)}\left(dkxd:x\ne0,x\ne5\right)\\ =\dfrac{3x-x-1}{x\left(x-5\right)}=\dfrac{2x-1}{x^2-5x}\)
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\(\dfrac{8\left(y+2\right)}{3x^2}.\dfrac{15x^5}{4\left(y+2\right)^2}\left(dkxd:x\ne0,y\ne-2\right)\\ =\dfrac{8}{4}.\dfrac{15x^2.x^3}{3x^2}=10x^3\)
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\(\dfrac{8\left(y-1\right)}{3x^2-3}:\dfrac{4\left(y-1\right)^3}{x^2-2x+1}\left(dkxd:x\ne1,x\ne-1\right)\\ =\dfrac{8\left(y-1\right)}{3\left(x-1\right)\left(x+1\right)}.\dfrac{\left(x-1\right)^2}{4\left(y-1\right)^3}\\ =\dfrac{2\left(x-1\right)}{3\left(x+1\right)\left(y-1\right)^2}\)
TT
21 tháng 1 2022
a) \(\dfrac{x+1}{4}-\dfrac{5+2x}{8}=\dfrac{3-4x}{2}\)
⇔\(\dfrac{2\left(x+1\right)}{8}-\dfrac{5+2x}{8}=\dfrac{4\left(3-4x\right)}{8}\)
⇔ 2x + 2 - 5 - 2x = 12 -16x
⇔ 16x = 15
⇔ x = 15/16
b) \(\dfrac{4-3x}{5}-\dfrac{4-x}{10}=\dfrac{x+2}{2}\)
⇔\(\dfrac{2\left(4-3x\right)}{10}-\dfrac{4-x}{10}=\dfrac{5\left(x+2\right)}{10}\)
⇔ 8 - 6x - 4 + x = 5x + 10
⇔ 10x = -6
⇔ x = -6/10
21 tháng 1 2022
Câu 1:
x + 1/4 - 5 + 2x/8 = 3 - 4x/2
<=> 2x + 2/8 - 5 + 2x/8 = 12 - 16x/8
<=> 2x + 2 - 5 - 2x = 12 - 16x
<=> -3 = 12 - 16x <=> 15 = 16x <=> x = 15/16
Câu 2:
4 - 3x/5 - 4 - x/10 = x + 2/2
<=> 8 - 6x/10 - 4 - x/10 = 5x + 10/10
<=> 8 - 6x - 4 + x = 5x + 10
<=> 4 - 5x = 5x + 10
<=> 4 = 10x + 10 <=> 10x = -6 <=> x = -3/5
H9
HT.Phong (9A5)
CTVHS
18 tháng 7 2023
1) \(3^x=\dfrac{9^8}{27^3\cdot81^2}\)
\(\Rightarrow3^x=\dfrac{\left(3^2\right)^8}{\left(3^3\right)^3\cdot\left(3^4\right)^2}\)
\(\Rightarrow3^x=\dfrac{3^{16}}{3^{15}}\)
\(\Rightarrow3^x=3\)
\(\Rightarrow x=1\)
2) \(\dfrac{2^{4-x}}{16^5}=32^6\)
\(\Rightarrow\dfrac{2^{4-x}}{\left(2^4\right)^5}=\left(2^5\right)^6\)
\(\Rightarrow\dfrac{2^{4-x}}{2^{20}}=2^{30}\)
\(\Rightarrow2^{4-x}=2^{20}\cdot2^{30}\)
\(\Rightarrow2^{4-x}=2^{50}\)
\(\Rightarrow4-x=50\)
\(\Rightarrow x=-46\)
3) \(\dfrac{2^{2x-3}}{4^{10}}=8^3\cdot16^5\)
\(\Rightarrow\dfrac{2^{2x-3}}{\left(2^2\right)^{10}}=\left(2^3\right)^3\cdot\left(2^4\right)^5\)
\(\Rightarrow\dfrac{2^{2x-3}}{2^{20}}=2^{29}\)
\(\Rightarrow2^{2x-3}=2^{49}\)
\(\Rightarrow2x-3=49\)
\(\Rightarrow2x=52\)
\(\Rightarrow x=26\)
AH
Akai Haruma
Giáo viên
13 tháng 12 2021
Lời giải:
a.
\(\frac{10}{x+2}=\frac{60}{6(x+2)}=\frac{60(x-2)}{6(x+2)(x-2)}=\frac{60(x-2)}{6(x^2-4)}\)
\(\frac{5}{2x-4}=\frac{15(x+2)}{6(x-2)(x+2)}=\frac{15(x+2)}{6(x^2-4)}\)
\(\frac{1}{6-3x}=\frac{x+2}{3(2-x)}=\frac{2(x+2)^2}{6(2-x)(2+x)}=\frac{-2(x+2)^2}{6(x^2-4)}\)
b.
\(\frac{1}{x+2}=\frac{x(2-x)}{x(x+2)(2-x)}=\frac{x(2-x)}{x(4-x^2)}\)
\(\frac{8}{2x-x^2}=\frac{8(x+2)}{(x+2)x(2-x)}=\frac{8(x+2)}{x(4-x^2)}\)
c.
\(\frac{4x^2-3x+5}{x^3-1}\)
\(\frac{1-2x}{x^2+x+1}=\frac{(1-2x)(x-1)}{(x-1)(x^2+x+1)}=\frac{-2x^2+3x-1}{x^3-1}\)
\(-2=\frac{-2(x^3-1)}{x^3-1}\)
26 tháng 7 2023
Tính phải k nhỉ?
`1)`
`2x + 3x + 5x`
`= (2 + 3 + 5)x`
`= 10x`
`2)`
`2.x - x + 3.x`
`= (2 - 1 + 3)x`
`= 4x`
`3)`
`9.x - 3 - 3.x`
`= (9 - 3)x - 3`
`= 6x - 3`
`4)`
Thiếu dấu, bạn bổ sung thêm
`5)`
`x - 0,2x - 0,1x`
`= (1 - 0,2 - 0,1)x`
`=0,7x`
`6)`
\(\dfrac{7}{2}x-\dfrac{1}{2}x=\left(\dfrac{7}{2}-\dfrac{1}{2}\right)x=3x\)
19 tháng 1 2022
Câu 1:
\(\left(x-2\right)\left(x^2+2x+4\right)+25x=x\left(x+5\right)\left(x-5\right)+8\)
\(\Leftrightarrow x^3-8+25x=x\left(x^2-25\right)+8\)
\(\Leftrightarrow x^3-8+25x=x^3-25x+8\)
\(\Leftrightarrow x^3-8+25x-x^3+25x-8=0\)
\(\Leftrightarrow50x-16=0\)
\(\Leftrightarrow50x=16\)
\(\Leftrightarrow x=\dfrac{8}{25}\)
B
19 tháng 1 2022
Câu 2 :
\(\dfrac{x+5}{4}+\dfrac{3+2x}{3}=\dfrac{6x-1}{3}-\dfrac{1-2x}{12}\)
<=> \(\dfrac{3\left(x+5\right)}{12}+\dfrac{4\left(3+2x\right)}{12}=\dfrac{4\left(6x-1\right)}{12}-\dfrac{1-2x}{12}\)
<=>\(\dfrac{3x+15+12+8x}{12}=\dfrac{24x-4-1+2x}{12}\)
<=> 3x + 15 + 12 + 8x = 24x - 4 - 1 +2x
<=> 11x+27 = 26x -5
<=> ( 26x - 5 ) - ( 11x + 27 ) = 0
<=> 15x - 32 = 0
<=> 15x = 32
<=> x = \(\dfrac{32}{15}\)
TV
1
NV
Nguyễn Việt Lâm
Giáo viên
11 tháng 12 2021
\(A=\dfrac{x^3-2x^2-15x}{x-5}=\dfrac{x\left(x^2-2x-15\right)}{x-5}=\dfrac{x\left(x+3\right)\left(x-5\right)}{x-5}=x\left(x+3\right)\)
\(A=x^2+3x=\left(x^2+3x+\dfrac{9}{4}\right)-\dfrac{9}{4}=\left(x+\dfrac{3}{2}\right)^2-\dfrac{9}{4}\ge-\dfrac{9}{4}\)
\(A_{min}=-\dfrac{9}{4}\)