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7(2x-1)-3(4x-1)=4(3x 2)-1; (5y 1) (1-y)= (9y 1) (1-3y); 20% (1-20%)(320-x)=320×40% 2(x-2) 2=x 1 2(x-2)-3(4x-1)=9(1-x) x/3 -5 = (5-x)/2 2(x 1) /3=5(x 1) /6 -1 (1/5)x 1 =(2x 1)/4 (5-2)/2 - (4 x)/3 =1 x/3 -1 = (1-x)/2 (x-2)/2 - (3x-2)/4 =-1 11x 64-2x=100-9x 15-(8-5x)=7x (4-3x) 3(x-7)-2[9-4(2-x)]=22 3/2[2/3(1/4x-1)-2]-x=2 2(x-2)-3(4x-1)=9(1-x) 11x 64-2x=100-9x 15-(8-5x)=7x (4-3x) 3(x-7)-2[9-4(2-x)]=22 3/2[2/3(1/4x-1)-2]-x=2 2(x-2) 2=x 1 7(2x-1)-3(4x-1)=4(3x 2)-1(5y 1) (1-y)= (9y 1) (1-3y)[ (- 2)-4 ]=x 220% (1-20%)(320-x)=320×40%2(x-2) 2=x 1 6。
2(x-2)-3(4x-1)=9(1-x) 7。11x 64-2x=100-9x 15-(8-5x)=7x (4-3x) 3(x-7)-2[9-4(2-x)]=22 3/2[2/3(1/4x-1)-2]-x=25x 1-2x=3x-23y-4=2y 187X*13=57Z/93=41 15X 863-65X=54 58Y*55=274892(x 2) 4=92(x 4)=103(x-5)=184x 8=2(x-1)3(x 3)=9 x6(x/2 1)=129(x 6)=632 x=2(x-1/2)8x 3(1-x)=-27 x-2(x-1)=1x/3 -5 = (5-x)/2 2(x 1) /3=5(x 1) /6 -1 (1/5)x 1 =(2x 1)/4 (5-2)/2 - (4 x)/3 =1 15x-8(5x 1。
2(x-2)-3(4x-1)=9(1-x) 7。11x 64-2x=100-9x 15-(8-5x)=7x (4-3x) 3(x-7)-2[9-4(2-x)]=22 3/2[2/3(1/4x-1)-2]-x=25x 1-2x=3x-23y-4=2y 187X*13=57Z/93=41 15X 863-65X=54 58Y*55=274892(x 2) 4=92(x 4)=103(x-5)=184x 8=2(x-1)3(x 3)=9 x6(x/2 1)=129(x 6)=632 x=2(x-1/2)8x 3(1-x)=-27 x-2(x-1)=1x/3 -5 = (5-x)/2 2(x 1) /3=5(x 1) /6 -1 (1/5)x 1 =(2x 1)/4 (5-2)/2 - (4 x)/3 =1 15x-8(5x 1。
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(7) 设 p = y', 则 y'' = dp/dx = (dp/dy)(dy/dx) = pdp/dy, 得
y^3 pdp/dy = 1, pdp = dy/y^3, (1/2)p^2 = (-1/2)y^(-2) + (1/2)C1,
p^2 = -1/y^2 + C1, y' = ±√(C1y^2-1)/y, ydy/√(C1y^2-1) = ±dx
d(C1y^2-1)/√(C1y^2-1) = ±2C1dx
2√(C1y^2-1) = ±2C1x + 2C2, √(C1y^2-1) = ±C1x + C2.
(8) 设 p = y', 则 y'' = dp/dx = (dp/dy)(dy/dx) = pdp/dy, 得
pdp/dy = p^3+p , p = 0, 或 dp/dy = p^2+1
dp/dy = p^2+1 即 dp/(1+p^2) = dy, arctanp = y + C1
y' = p = tan(y+C1), dy/tan(y+C1) = dx,
dsin(y+C1)/sin(y+C1) = dx, ln[sin(y+C1)] = x + lnC2
sin(y+C1) = C2e^x.
p = 0 即 y' = 0, y = C3
通解是 sin(y+C1) = C2e^x 或 y = C3。
y^3 pdp/dy = 1, pdp = dy/y^3, (1/2)p^2 = (-1/2)y^(-2) + (1/2)C1,
p^2 = -1/y^2 + C1, y' = ±√(C1y^2-1)/y, ydy/√(C1y^2-1) = ±dx
d(C1y^2-1)/√(C1y^2-1) = ±2C1dx
2√(C1y^2-1) = ±2C1x + 2C2, √(C1y^2-1) = ±C1x + C2.
(8) 设 p = y', 则 y'' = dp/dx = (dp/dy)(dy/dx) = pdp/dy, 得
pdp/dy = p^3+p , p = 0, 或 dp/dy = p^2+1
dp/dy = p^2+1 即 dp/(1+p^2) = dy, arctanp = y + C1
y' = p = tan(y+C1), dy/tan(y+C1) = dx,
dsin(y+C1)/sin(y+C1) = dx, ln[sin(y+C1)] = x + lnC2
sin(y+C1) = C2e^x.
p = 0 即 y' = 0, y = C3
通解是 sin(y+C1) = C2e^x 或 y = C3。
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