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1: Ta có: \(4x^2-36=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
2: Ta có: \(\left(x-1\right)^2+x\left(4-x\right)=11\)
\(\Leftrightarrow x^2-2x+1+4x-x^2=11\)
\(\Leftrightarrow2x=10\)
hay x=5
bài 11
a) \(x^2-xy+x\\ =x\left(x-y+1\right)\)
b)
\(x^2-2xy-4+y^2\\ =\left(x^2-2xy+y^2\right)-4\\ =\left(x-y\right)^2-4\\ =\left(x-y-2\right)\left(x-y+2\right)\)
c)
\(x^3-x^2-16x+16\\ =x^2\left(x-1\right)-16\left(x-1\right)\\ =\left(x-1\right)\left(x-4\right)\left(x+4\right)\)
bài 12
\(2x\left(x-5\right)-x\left(3+2x\right)=26\)
\(2x^2-10x-3x-2x^2=26\)
\(-13x=26\\ x=-2\)
b)
\(2\left(x+5\right)-x^2-5x=0\\ 2\left(x+5\right)-x\left(x+5\right)=0\\ \left(x+5\right)\left(2-x\right)=0\\ \left[{}\begin{matrix}x+5=0\\2-x=0\end{matrix}\right.\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
\(8,1-\left(x-6\right)=4\left(2-2x\right)\)
\(\Leftrightarrow1-x+6=8-8x\)
\(\Leftrightarrow-x+8x=8-1-6\)
\(\Leftrightarrow7x=1\)
\(\Leftrightarrow x=\dfrac{1}{7}\)
\(9,\left(3x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-5\end{matrix}\right.\)
\(10,\left(x+3\right)\left(x^2+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^2+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\varnothing\end{matrix}\right.\)
`8)1-(x-5)=4(2-2x)`
`<=>1-x+5=8-6x`
`<=>5x=2<=>x=2/5`
`9)(3x-2)(x+5)=0`
`<=>[(x=2/3),(x=-5):}`
`10)(x+3)(x^2+2)=0`
Mà `x^2+2 > 0 AA x`
`=>x+3=0`
`<=>x=-3`
`11)(5x-1)(x^2-9)=0`
`<=>(5x-1)(x-3)(x+3)=0`
`<=>[(x=1/5),(x=3),(x=-3):}`
`12)x(x-3)+3(x-3)=0`
`<=>(x-3)(x+3)=0`
`<=>[(x=3),(x=-3):}`
`13)x(x-5)-4x+20=0`
`<=>x(x-5)-4(x-5)=0`
`<=>(x-5)(x-4)=0`
`<=>[(x=5),(x=4):}`
`14)x^2+4x-5=0`
`<=>x^2+5x-x-5=0`
`<=>(x+5)(x-1)=0`
`<=>[(x=-5),(x=1):}`
a: Sửa đề: \(\left(2x^2-3x-1\right)^2-3\left(2x^2-3x-5\right)-16=0\)
\(\Leftrightarrow\left(2x^2-3x-1\right)^2-3\left(2x^2-3x-1-4\right)-16=0\)
\(\Leftrightarrow\left(2x^2-3x-1\right)^2-3\left(2x^2-3x-1\right)-4=0\)
\(\Leftrightarrow\left(2x^2-3x-1-4\right)\left(2x^2-3x-1+1\right)=0\)
\(\Leftrightarrow\left(2x^2-3x-5\right)\left(2x^2-3x\right)=0\)
\(\Leftrightarrow\left(2x^2-5x+2x-5\right)\cdot x\cdot\left(2x-3\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(x+1\right)x\left(2x-3\right)=0\)
hay \(x\in\left\{\dfrac{5}{2};-1;0;\dfrac{3}{2}\right\}\)
b: \(\Leftrightarrow\left(x^2+x\right)^2+4\left(x^2+x\right)-12=0\)
\(\Leftrightarrow\left(x^2+x+6\right)\left(x^2+x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)=0\)
hay \(x\in\left\{-2;1\right\}\)
(2x + 5 )2 + (4x+10)(3-x) + x2-6x+9=0
=>(2x+5)2_ 2(2x+5)(x-3) + (x-3)2=0
=>[(2x+5)-(x-3)]2 =0
=>(x+8)2=0
=> x+8=0
=> x=-8
x2 - 5x - 36 = 0
=> x2 - 9x + 4x - 36 = 0
=> x(x - 9) + 4(x - 7) = 0
=> (x + 4)(x - 7) = 0
=> \(\orbr{\begin{cases}x+4=0\\x-7=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=-4\\x=7\end{cases}}\)
6x2 - (2x + 5)(3x - 2) = -12
=> 6x2 - 6x2 + 4x - 15x + 10 = -12
=> -11x = -22
=> x = 2
x2 - 25 = 6x - 9
=> x2 - 25 - 6x + 9 = 0
=> x2 - 6x - 16 = 0
=> x2 - 8x + 2x - 16 = 0
=> x(x - 8) + 2(x - 8) = 0
=> (x + 2)(x - 8) = 0
=> \(\orbr{\begin{cases}x+2=0\\x-8=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=-2\\x=8\end{cases}}\)
\(\left(x-2\right)^3-\left(x+5\right)\left(x^2-5x+25\right)+6x^2=11\)
=>\(x^3-6x^2+12x-8-\left(x^3+125\right)+6x^2=11\)
=>\(x^3+12x-8-x^3-125=11\)
=>12x-133=11
=>12x=144
=>\(x=\dfrac{144}{12}=12\)
\(\Leftrightarrow30xy-72x+55y-132-42x+16=0\)
\(\Leftrightarrow30xy-72x+55y-42x=0-16+132\)
........................................ Bạn tự làm tiếp nhé!!!