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- a, \(\frac{2}{3}x+\frac{5}{7}=\frac{3}{10}\Rightarrow\frac{2}{3}x=\frac{3}{10}-\frac{5}{7}=\frac{-29}{70}\Rightarrow x=\frac{-29}{70}:\frac{2}{3}=\frac{-87}{140}\)
- b, \(\frac{3}{4}x-\frac{1}{8}=\frac{3}{7}\Rightarrow\frac{3}{4}x=\frac{3}{7}+\frac{1}{8}=\frac{31}{56}\Rightarrow x=\frac{31}{56}:\frac{3}{4}=\frac{31}{42}\)
c, \(\frac{-21}{13}x+\frac{1}{3}=\frac{2}{3}\Rightarrow\frac{-21}{13}x=\frac{2}{3}-\frac{1}{3}=\frac{1}{3}\Rightarrow x=\frac{1}{3}:\frac{-21}{3}=\frac{-1}{21}\)
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b)\(\frac{3}{4}x-\frac{1}{8}=\frac{3}{7}\)\(\Leftrightarrow\frac{3}{4}x=\frac{3}{7}+\frac{1}{8}=\frac{31}{56}\)\(\Leftrightarrow x=\frac{31}{56}:\frac{3}{4}=\frac{31}{42}\)
c)\(-\frac{21}{13}x+\frac{1}{3}=\frac{2}{3}\Leftrightarrow-\frac{21}{13}x=\frac{2}{3}-\frac{1}{3}=\frac{1}{3}\Leftrightarrow x=\frac{1}{3}:-\frac{21}{13}=-\frac{13}{63}\)
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a) \(5-\frac{2x}{3}=4x-\frac{1}{-5}\)
\(\frac{75-10x}{15}=\frac{60x+3}{15}\)
75 - 10x = 60x +3
72 = 70x
\(\frac{72}{70}\) = x
x =\(\frac{36}{35}\)
Vậy x = \(\frac{36}{35}\)
b) \(2x-\frac{10}{6}=\frac{-27}{5}-x\)
\(2x-\frac{5}{3}=\frac{-27}{5}-x\)
\(\frac{30x-25}{15}=\frac{-81-15}{15}\)
30x =-96+25
30x =-71
x= -71/30
Vậy x= -71/30
c) \(13x-\frac{2}{2x}+5=\frac{76}{17}\)
13x - 1/x +5 = 76/17
\(\frac{221x-17+85}{17x}=\frac{76x}{17x}\)
221x +68 = 76x
221x-76x =-68
145x =-68
x =\(\frac{-68}{145}\)
Vậy .........
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\(\frac{5-2x}{3}=\frac{4x-1}{-5}\)
-5(5-2x) = 3(4x-1)
-25 + 10x = 12x - 3
10x - 12x = -3 + 25
-2x = 22
x= -11
Nhân chéo như trên rồi tự làm nha
Học tốt~
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c: \(\dfrac{4x^3-8x^2+13x-5}{2x-1}=\dfrac{4x^3-2x^2-6x^2+3x+10x-5}{2x-1}\)
=2x^2-3x+5
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a: \(=-2x^2\cdot3x+2x^2\cdot4X^3-2x^2\cdot7+2x^2\cdot x^2\)
\(=8x^5+2x^4-6x^3-14x^2\)
b: \(=2x^3-3x^2-5x+6x^2-9x-15\)
\(=2x^3+3x^2-14x-15\)
c: \(=\dfrac{-6x^5}{3x^3}+\dfrac{7x^4}{3x^3}-\dfrac{6x^3}{3x^3}=-2x^2+\dfrac{7}{3}x-2\)
d: \(=\dfrac{\left(3x-2\right)\left(3x+2\right)}{3x+2}=3x-2\)
e: \(=\dfrac{2x^4-8x^3-6x^2-5x^3+20x^2+15x+x^2-4x-3}{x^2-4x-3}\)
=2x^2-5x+1
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\(H\left(x\right)=9x^4-3x^3-11x^2-7x+12\)
\(K\left(x\right)=-8x^4+10x^3+4x^2-7x-12\)
\(A\left(x\right)=H\left(x\right)-K\left(x\right)\)
\(=17x^4-10x^3-15x^2+24\)
Để \(A\left(x\right)=x^4-13x^3-14x^2\) nên \(17x^4-10x^3-15x^2+24=x^4-13x^3-14x^2\)
\(\Leftrightarrow16x^4+3x^3-x^2+24=0\)
Đến đây mình bí rồi, xin lỗi bạn!
\(=\dfrac{-x^3+4x^2-13x+10}{x-5}\)
\(=\dfrac{-x^3+5x^2-x^2+5x-18x+90-80}{x-5}\)
\(=-x^2-x-18-\dfrac{80}{x-5}\)
HÌNH NHƯ sai hay sao á =()?