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c: Ta có: \(\left(2x-3\right)^2-\left(2x-3\right)\left(x-10\right)=7\)
\(\Leftrightarrow4x^2-12x+9-2x^2+20x+3x-30=7\)
\(\Leftrightarrow11x=28\)
hay \(x=\dfrac{28}{11}\)
d: Ta có: \(\left(3x-4\right)^2-9\left(x-3\right)\left(x+3\right)=8\)
\(\Leftrightarrow9x^2-24x+16-9x^2+81=8\)
\(\Leftrightarrow-24x=-89\)
hay \(x=\dfrac{89}{24}\)
f: Ta có: \(\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=16\)
\(\Leftrightarrow x^2+8x+16-x^2+1=16\)
\(\Leftrightarrow8x=-1\)
hay \(x=-\dfrac{1}{8}\)
`a,x(x-1)-(x+2)^2=1`
`<=>x^2-x-x^2-4x-4=1`
`<=>-5x=5`
`<=>x=-1`
`b,(x+5)(x-3)-(x-2)^2=-1`
`<=>x^2+2x-15-x^2+4x-4+1=0`
`<=>6x-18=0`
`<=>x-3=0`
`<=>x=3`
`c,x(2x-4)-(x-2)(2x+3)=0`
`<=>2x(x-2)-(x-2)(2x+3)=0`
`<=>(x-2)(2x-2x-3)=0`
`<=>-3(x-2)=0`
`<=>x-2=0`
`<=>x=2`
`d,x(3x+2)+(x+1)^2-(2x-5)(2x+5)=-12`
`<=>3x^2+2x+x^2+2x+1-4x^2+25=-12`
`<=>4x+26=-12`
`<=>4x=-38`
`<=>x=-19/2`
a: Ta có: \(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=6\)
\(\Leftrightarrow x^2+5x+6-x^2-3x+10=6\)
\(\Leftrightarrow2x=-10\)
hay x=-5
b: Ta có: \(\left(3x+2\right)\left(2x+9\right)-\left(x+2\right)\left(6x+1\right)=\left(x+1\right)-\left(x-6\right)\)
\(\Leftrightarrow6x^2+27x+4x+18-6x^2-x-12x-2=x+1-x+6\)
\(\Leftrightarrow18x+16=7\)
hay \(x=-\dfrac{1}{2}\)
c: Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)
\(\Leftrightarrow3\left(6x^2-2x-3x+1\right)-\left(18x^2-2x-27x+3\right)=0\)
\(\Leftrightarrow18x^2-15x+3-18x^2+27x-3=0\)
hay x=0
a) \(\left(x+1\right)^3-\left(x-1\right)^3-6\cdot\left(x-1\right)^2=10\)
\(\Rightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\cdot\left(x^2-2x+1\right)=10\)
\(\Rightarrow6x^2+2-6x^2+12x-6=10\)
\(\Rightarrow12x-4=10\)
\(\Rightarrow12x=14\)
\(\Rightarrow x=\dfrac{7}{6}\)
b) \(x\left(x+5\right)\left(x-5\right)-\left(x+2\right)\left(x^2-2x+4\right)=42\)
\(\Rightarrow x\left(x^2-25\right)-\left(x^3+8\right)=42\)
\(\Rightarrow x^3-25x-x^3-8=42\)
\(\Rightarrow-25x-8=42\)
\(\Rightarrow-25x=50\)
\(\Rightarrow x=\dfrac{50}{-25}=-2\)
c) \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=49\)
\(\Rightarrow x^3-6x^2+12x-8-\left(x^3-27\right)+6\left(x^2+2x+1\right)=49\)
\(\Rightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=49\)
\(\Rightarrow24x+25=49\)
\(\Rightarrow24x=24\)
\(\Rightarrow x=\dfrac{24}{24}=1\)
a)\(=\left(x^2+2x+1\right)-y^2=\left(x+1\right)^2-y^2=\left(x+1+y\right)\left(x+1-y\right)\)
b)\(=\left(x+9\right)^2-\left(6x\right)^2=\left(x+9-6x\right)\left(x+9+6x\right)=\left(-5x+9\right)\left(7x+9\right)\)
c)\(=\left(x^2-2xy+y^2\right)-\left(z^2-2zt+t^2\right)=\left(x-y\right)^2-\left(z-t\right)^2\\ =\left(x-y+z-t\right)\left(x-y-z+t\right)\)
a: =>x^2-25-x^2-3x=10
=>-3x=35
=>x=-35/3
b: =>4x^2-9-4(x^2+4x+4)=5
=>4x^2-9-4x^2-16x-16-5=0
=>-16x-30=0
=>x=-15/8
c: =>9x^2+45x-9x^2+4=7
=>45x=3
=>x=1/15
d: =>x^3+3x^2+3x+1-x^3-3x^2+5x=8
=>8x=7
=>x=7/8
\(a,=x^2-4-x^2+2x+3=2x-1\\ b,=x^3+3x^2-5x-15+x^2-x^3+4x-4x^2=-x-15\\ c,=2x^2+3x-10x-15-2x^2+6x+x+7=-8\\ d,=\left(2x+1+3x-1\right)^2=25x^2\)
\(a)\dfrac{x-3}{x-2}+\dfrac{x-2}{x-4}=-1.\left(x\ne2;4\right).\\ \Leftrightarrow\dfrac{\left(x-3\right)\left(x-4\right)+\left(x-2\right)^2}{\left(x-2\right)\left(x-4\right)}=-1.\\ \Rightarrow x^2-4x-3x+12+x^2-4x+4+x^2-4x-2x+8=0.\\ \Leftrightarrow3x^2-17x+24=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}.\\x=3.\end{matrix}\right.\) (TM).
\(b)3x+12=0.\\ \Leftrightarrow3x=-12.\\ \Leftrightarrow x=-4.\)
\(c)5+2x=x-5.\\ \Leftrightarrow2x-x=-5-5.\\ \Leftrightarrow x=-10.\)
\(d)2x\left(x-2\right)+5\left(x-2\right)=0.\\ \Leftrightarrow\left(2x+5\right)\left(x-2\right)=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5}{2}.\\x=2.\end{matrix}\right.\)
\(e)\dfrac{3x-4}{2}=\dfrac{4x+1}{3}.\\ \Rightarrow3\left(3x-4\right)-2\left(4x+1\right)=0.\\ \Leftrightarrow9x-12-8x-2=0.\\ \Leftrightarrow x=14.\)
\(f)\dfrac{2x}{x-1}-\dfrac{x}{x+1}=1.\left(x\ne\pm1\right).\\ \Leftrightarrow\dfrac{2x^2+2x-x^2+x}{x^2-1}=1.\\ \Leftrightarrow x^2+3x-x^2+1=0.\\ \Leftrightarrow3x+1=0.\\ \Leftrightarrow x=\dfrac{-1}{3}.\)
\(g)\dfrac{2x}{x-1}+\dfrac{3-2x}{x+2}=\dfrac{6}{\left(x-1\right)\left(x+2\right)}.\left(x\ne1;-2\right).\\ \Leftrightarrow\dfrac{2x^2+4x+\left(3-2x\right)\left(x-1\right)}{\left(x-1\right)\left(x+2\right)}=\dfrac{6}{\left(x-1\right)\left(x+2\right)}.\\ \Rightarrow2x^2+4x+3x-3-2x^2+2x-6=0.\\ \Leftrightarrow9x=9.\)
\(\Leftrightarrow x=1\left(koTM\right).\)
a) (x+3)^2 - (2-x)^2 = 1
x^2 + 6x + 9 - (4 - 4x + x^2) = 1
x^2 + 6x + 9 - 4 + 4x - x^2 = 1
10x + 5 = 1
10x = -4
x = -4/10
x = -2/5
Vậy giá trị của x là -2/5.
b) 5(x-2)^2 - (x+3)^2 = (2x-1)^2
5(x^2 - 4x + 4) - (x^2 + 6x + 9) = 4x^2 - 4x + 1
5x^2 - 20x + 20 - x^2 - 6x - 9 = 4x^2 - 4x + 1
4x^2 - 26x + 30 = 4x^2 - 4x + 1
-26x + 30 = -4x + 1
-22x = -29
x = 29/22
Vậy giá trị của x là 29/22.
c) (x-1)^2 - x(x+5)^2 = 7
x^2 - 2x + 1 - x(x^2 + 10x + 25) = 7
x^2 - 2x + 1 - x^3 - 10x^2 - 25x = 7
-x^3 - 9x^2 - 27x - 6 = 0
d) (3x-2)^2 - 9(x+2)^2 = 3
9x^2 - 12x + 4 - 9x^2 - 36x - 36 = 3
-48x - 32 = 3
-48x = 35
x = -35/48
Vậy giá trị của x là -35/48.