Giúp mik với!
\(\frac{1}{2}x\frac{1}{4}x^2\frac{x^3}{8}2y4y^28y^3\)
Những câu hỏi liên quan
\(\frac{1}{2}x\frac{1}{4}x^2\frac{x^3}{8}2y4y^28y^3\)
\(\frac{1}{2}x\frac{1}{4}x^2\frac{x^3}{8}2y4y^28y^3\)
Thu gọn các đơn thức sau:a) 1frac{1}{4}x2y left(frac{-5}{6}xyright)^0left(-2frac{1}{3}xyright)b) frac{1}{2}xfrac{1}{4}x^2frac{x^3}{8}2y4y^28y^3c) left(2frac{1}{3}x^2y^3right)^{10}left(frac{3}{7}x^5y^4right)^{10}d) left(-frac{a}{2}right)^33xyleft(4a^2x^3right)left(4frac{1}{3}ay^2right)
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Thu gọn các đơn thức sau:
a) 1\(\frac{1}{4}\)x2y \(\left(\frac{-5}{6}xy\right)^0\)\(\left(-2\frac{1}{3}xy\right)\)
b) \(\frac{1}{2}x\frac{1}{4}x^2\frac{x^3}{8}2y4y^28y^3\)
c) \(\left(2\frac{1}{3}x^2y^3\right)^{10}\left(\frac{3}{7}x^5y^4\right)^{10}\)
d) \(\left(-\frac{a}{2}\right)^33xy\left(4a^2x^3\right)\left(4\frac{1}{3}ay^2\right)\)
a.\(\frac{5}{4}x^2y.\left(\frac{-5}{6}xy\right)^0\left(\frac{-7}{3}xy\right)\)= \(\frac{5}{4}x^2y.1.\left(\frac{-7}{3}xy\right)\)= \(\frac{-35}{12}x^3.y^2\)
câu b, c,d làm tương tự như trên nha ^.^
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Thu gọn các đơn thức sau rồi cho biết hệ số, phần biến, bậc của mỗi đơn thức:a) frac{1}{2}xfrac{1}{4}x^2^{ }frac{x^3}{8}2y4y^28y^3b)left(2frac{1}{3}x^2y^3right)^{10}left(frac{3}{7}x^5y^4right)^{10}c)left(-frac{2}{3}a^2bright)^3ba^2left[-frac{9}{4}left(ab^2right)^3right]left(frac{a^3b}{2}right)^2d)left(-frac{1}{2}afrac{4}{3}a^26aright)^3.3^3left(-b^2right)frac{1}{6}b8^3e)left(-frac{a}{2}right)^33xyleft(4a^2x^3right)left(4frac{1}{3}ay^2right) (a là hằng số khác 0)
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Thu gọn các đơn thức sau rồi cho biết hệ số, phần biến, bậc của mỗi đơn thức:
a) \(\frac{1}{2}x\frac{1}{4}x^2^{ }\frac{x^3}{8}2y4y^28y^3\)
b)\(\left(2\frac{1}{3}x^2y^3\right)^{10}\left(\frac{3}{7}x^5y^4\right)^{10}\)
c)\(\left(-\frac{2}{3}a^2b\right)^3ba^2\left[-\frac{9}{4}\left(ab^2\right)^3\right]\left(\frac{a^3b}{2}\right)^2\)
d)\(\left(-\frac{1}{2}a\frac{4}{3}a^26a\right)^3.3^3\left(-b^2\right)\frac{1}{6}b8^3\)
e)\(\left(-\frac{a}{2}\right)^33xy\left(4a^2x^3\right)\left(4\frac{1}{3}ay^2\right)\) (a là hằng số khác 0)
Bài 1:
a, x+1/2=3/8.4/5
b, 5/16:x-1/4=5/8
c,-1/4.x+3/7.x=2
d, 22/9-(x+1/2)^2=7/3
e, \(\frac{1}{4}-x+5\frac{1}{8}=6\frac{1}{8}\)
Giúp mik với, đang gấp lắm!!!!!!!!!!!!!!!!!!!!!!!!!!!!
giúp mik vs mai mik kiểm tra rùi
a) $\frac{x-1}{x}$ - $\frac{1}{x+1}$ = $\frac{2x-1}{x2+x}$
b) (x+2).(5-3x)=0
c)$\frac{5(1-2x)}{3}$ + $\frac{x}{2}$ = $\frac{3(x-5)}{4}$ - 2
d)$(x+2)^{2}$ - (x-1).(x+3) = (2x-4).(x+4)-3
e)$(2x-3)^{2}$ = (2x-3).(x+1)
a:=>x^2-1-x=2x-1
=>x^2-x-1=2x-1
=>x^2-3x=0
=>x=0(loại) hoặc x=3(nhận)
b:=>x+2=0 hoặc 5-3x=0
=>x=-2 hoặc x=5/3
c:=>20(1-2x)+6x=9(x-5)-24
=>20-40x+6x=9x-45-24
=>-34x+20=9x-69
=>-43x=-89
=>x=89/43
d: =>x^2+4x+4-x^2-2x+3=2x^2+8x-4x-16-3
=>2x^2+4x-19=-2x+7
=>2x^2+6x-26=0
=>x^2+3x-13=0
=>\(x=\dfrac{-3\pm\sqrt{61}}{2}\)
e: =>(2x-3)(2x-3-x-1)=0
=>(2x-3)(x-4)=0
=>x=4 hoặc x=3/2
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\(\frac{x^2+4}{8}+\frac{x^2+3}{7}+\frac{x^2+2}{6}=\frac{x^2+1}{5}+\frac{x^2}{4}+\frac{x^2-1}{3}\)
GIÚP MÌNH VỚI
\(\Leftrightarrow\frac{x^2+4}{8}-1+\frac{x^2+3}{7}-1+\frac{x^2+2}{6}-1=\frac{x^2+1}{5}-1+\frac{x^2}{4}-1+\frac{x^2-1}{3}-1\)
\(\Leftrightarrow\frac{x^2-4}{8}+\frac{x^2-4}{7}+\frac{x^2-4}{6}-\frac{x^2-4}{5}-\frac{x^2-4}{4}-\frac{x^2-4}{3}=0\)
\(\Leftrightarrow\left(x^2-4\right)\left(\frac{1}{8}+\frac{1}{7}+\frac{1}{6}+\frac{1}{5}+\frac{1}{4}+\frac{1}{3}\right)\)
\(\Leftrightarrow x^2-4=0\Leftrightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)
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Giải phương trình:1) frac{7x-1}{6}+2xfrac{16-x}{5}2) frac{10x+3}{12}1+frac{6+8x}{9}3)frac{8x-3}{4}-frac{3x-2}{2}frac{2x-1}{2}+frac{x+3}{4}4)frac{3left(3-xright)}{8}+frac{2left(5-xright)}{3}frac{1-x}{2}-25)frac{2left(x-3right)}{7}+frac{x-5}{3}-frac{13x+4}{21}06)frac{6x+5}{2}-left(2x+frac{2x+1}{2}right)frac{10x+3}{4}7)frac{2x-1}{5}-frac{x-2}{3}-frac{x+7}{15}08)frac{x+4}{5}-x+4frac{x}{3}-frac{x-2}{2}giải giúp mik với ạ
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Giải phương trình:
1) \(\frac{7x-1}{6}+2x=\frac{16-x}{5}\)
2) \(\frac{10x+3}{12}=1+\frac{6+8x}{9}\)
3)\(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)
4)\(\frac{3\left(3-x\right)}{8}+\frac{2\left(5-x\right)}{3}=\frac{1-x}{2}-2\)
5)\(\frac{2\left(x-3\right)}{7}+\frac{x-5}{3}-\frac{13x+4}{21}=0\)
6)\(\frac{6x+5}{2}-\left(2x+\frac{2x+1}{2}\right)=\frac{10x+3}{4}\)
7)\(\frac{2x-1}{5}-\frac{x-2}{3}-\frac{x+7}{15}=0\)
8)\(\frac{x+4}{5}-x+4=\frac{x}{3}-\frac{x-2}{2}\)
giải giúp mik với ạ
1. \(\frac{7x-1}{6}+2x=\frac{16-x}{5}\)
\(\Leftrightarrow5\left(7x-1\right)+60x=6\left(16-x\right)\)
\(\Leftrightarrow35x-5+60x=96-6x\)
\(\Leftrightarrow95x-5=96-6x\)
\(\Leftrightarrow95x+6x=96+5\)
\(\Leftrightarrow101x=101\)
\(\Leftrightarrow x=1\)
2. \(\frac{10x+3}{12}=1+\frac{6+8x}{9}\)
\(\Leftrightarrow3\left(10x+3\right)=36+4\left(6+8x\right)\)
\(\Leftrightarrow30x+9=36+24+32x\)
\(\Leftrightarrow30x+9=32x+60\)
\(\Leftrightarrow30x-32x=60-9\)
\(\Leftrightarrow-2x=51\)
\(\Leftrightarrow x=-\frac{51}{2}\)
3. \(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)
\(\Leftrightarrow8x-3-2\left(3x-2\right)=2\left(2x-1\right)+x+3\)
\(\Leftrightarrow8x-3-6x+4=4x-2+x+3\)
\(\Leftrightarrow2x+1=5x+1\)
\(\Leftrightarrow2x=5x\)
\(\Leftrightarrow x=0\)
4) \(\frac{3\left(3-x\right)}{8}+\frac{2\left(5-x\right)}{3}=\frac{1-x}{2}-2\)
=> \(\frac{9-3x}{8}+\frac{10-2x}{3}=\frac{1-x}{2}-\frac{2}{1}\)
=> \(\frac{3\left(9-3x\right)}{24}+\frac{8\left(10-2x\right)}{24}=\frac{12\left(1-x\right)}{24}-\frac{48}{24}\)
=> \(\frac{27-9x}{24}+\frac{80-16x}{24}=\frac{12-12x}{24}-\frac{48}{24}\)
=> \(\frac{27-9x+80-16x}{24}=\frac{12-12x-48}{24}\)
=> 27 - 9x + 80 - 16x = 12 - 12x - 48
=> 27 - 9x + 80 - 16x - 12 + 12x + 48 = 0
=> (27 + 80 - 12 + 48) + (-9x - 16x + 12x) = 0
=> 143 - 13x = 0
=> 13x = 143
=> x = 11
5) \(\frac{2\left(x-3\right)}{7}+\frac{x-5}{3}-\frac{13x+4}{21}=0\)
=> \(\frac{2x-6}{7}+\frac{x-5}{3}-\frac{13x+4}{21}=0\)
=> \(\frac{3\left(2x-6\right)}{21}+\frac{7\left(x-5\right)}{21}-\frac{13x+4}{21}=0\)
=> \(\frac{6x-18}{21}+\frac{7x-35}{21}-\frac{13x+4}{21}=0\)
=> \(\frac{6x-18+7x-35-13x-4}{21}=0\)
=> 6x - 18 + 7x - 35 - 13x - 4 = 0
=> (6x + 7x - 13x) + (-18 - 35 - 4) = 0
=> -57 = 0(vô nghiệm)
6) \(\frac{6x+5}{2}-\left(2x+\frac{2x+1}{2}\right)=\frac{10x+3}{4}\)
=> \(\frac{6x+5}{2}-\frac{10x+3}{4}=2x+\frac{2x+1}{2}\)
=> \(\frac{2\left(6x+5\right)}{4}-\frac{10x+3}{4}=\frac{8x}{4}+\frac{2\left(2x+1\right)}{4}\)
=> \(\frac{12x+10}{4}-\frac{10x+3}{4}=\frac{8x}{4}+\frac{4x+2}{4}\)
=> \(\frac{12x+10-\left(10x+3\right)}{4}=\frac{8x+4x+2}{4}\)
=> \(\frac{12x+10-10x-3}{4}=\frac{12x+2}{4}\)
=> \(12x+10-10x-3=12x+2\)
=> \(2x+10-3=12x+2\)
=> 2x + 10 - 3 - 12x - 2 = 0
=> (2x - 12x) + (10 - 3 - 2) = 0
=> -10x + 5 = 0
=> -10x = -5
=> x = 1/2
7) \(\frac{2x-1}{5}-\frac{x-2}{3}-\frac{x+7}{15}=0\)
=> \(\frac{3\left(2x-1\right)}{15}-\frac{5\left(x-2\right)}{15}-\frac{x+7}{15}=0\)
=> \(\frac{6x-3}{15}-\frac{5x-10}{15}-\frac{x+7}{15}=0\)
=> \(\frac{6x-3-\left(5x-10\right)-\left(x+7\right)}{15}=0\)
=> 6x - 3 - 5x + 10 - x - 7 = 0
=> (6x - 5x - x) + (-3 + 10 - 7) = 0
=> 0x + 0 = 0
=> 0x = 0
=> x tùy ý
Bài 8 tự làm nhé
mai mik kiểm tra rùi giúp mik vs pls
a) $\frac{x-1}{x}$ - $\frac{1}{x+1}$ = $\frac{2x-1}{x2+x}$
b) (x+2).(5-3x)=0
c)$\frac{5(1-2x)}{3}$ + $\frac{x}{2}$ = $\frac{3(x-5)}{4}$ - 2
\(\dfrac{x-1}{x}-\dfrac{1}{x+1}=\dfrac{2x-1}{x^2+x}\)
\(\Leftrightarrow\dfrac{x-1}{x}-\dfrac{1}{x+1}=\dfrac{2x-1}{x\left(x+1\right)}\)
ĐKXĐ : \(\left\{{}\begin{matrix}x\ne0\\x+1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne-1\end{matrix}\right.\)
Ta có : `(x-1)/x -1/(x+1) =(2x-1)/(x(x+1))`
\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x+1\right)}{x\left(x+1\right)}-\dfrac{x}{x\left(x+1\right)}=\dfrac{2x-1}{x\left(x+1\right)}\)
`=> x^2 +x -x-1 -x-2x+1=0`
`<=> x^2 -3x =0`
`<=> x(x-3)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=3\end{matrix}\right.\)
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`(x+2)(5-3x)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\5-3x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\3x=5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{5}{3}\end{matrix}\right.\)
__
\(\dfrac{5\left(1-2x\right)}{3}+\dfrac{x}{2}=\dfrac{3\left(x-5\right)}{4}-2\)
\(\Leftrightarrow\dfrac{20\left(1-2x\right)}{12}+\dfrac{6x}{12}=\dfrac{9\left(x-5\right)}{12}-\dfrac{24}{12}\)
`<=> 2x- 40x + 6x = 9x - 45 -24`
`<=> 2x- 40x + 6x-9x + 45 +24=0`
`<=>-41x+69=0`
`<=>-41x=-69`
`<=> x=69/41`
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a:=>x^2-1-x=2x-1
=>x^2-x-1=2x-1
=>x^2-3x=0
=>x=0(loại) hoặc x=3(nhận)
b:=>x+2=0 hoặc 5-3x=0
=>x=-2 hoặc x=5/3
c:=>20(1-2x)+6x=9(x-5)-24
=>20-40x+6x=9x-45-24
=>-34x+20=9x-69
=>-43x=-89
=>x=89/43
d: =>x^2+4x+4-x^2-2x+3=2x^2+8x-4x-16-3
=>2x^2+4x-19=-2x+7
=>2x^2+6x-26=0
=>x^2+3x-13=0
=>\(x=\dfrac{-3\pm\sqrt{61}}{2}\)
e: =>(2x-3)(2x-3-x-1)=0
=>(2x-3)(x-4)=0
=>x=4 hoặc x=3/2
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