\(\left(x^2+5x\right)+10\left(x^2-5x\right)+24=0\)
\(\Leftrightarrow\left(x^2+5x\right)-10\left(x^2+5x\right)+24=0\)
\(\Leftrightarrow\left(x^2+5x\right)\left(1-10\right)+14=0\)
\(\Leftrightarrow\left(-9\right)\left(x^2+5x\right)+14=0\)
\(\Leftrightarrow-9\left(x^2+5x\right)=-14\)
\(\Leftrightarrow x^2+5x=\frac{14}{9}\)
\(\Leftrightarrow x=0,2938.....\)
b) \(-2\left(x-1\right)^2=0\) => x = 1
(x-4)(x-5)(x-6)(x-7)=1680
=>[(x-4)(x-7)][(x-5)(x-6)]=1680
=>(x2 - 11x+28)(x2 - 11x+30)=1680
=>(x2 - 11x+28)(x2 - 11x+28+2)=1680
dat t=x2 - 11x+28 ta dc t(t+2)=1680
=>t2+ 2t=1680=>t2 +2t- 1680=0
=>t2 +42t-40t -1680=0
=>t(t +42)-40(t +42)=0
=>(t -40)(t +42)=0
=>t +42=0 => t={-42;40}
t -40=0
vi t=x2 - 11x+28 nen x2 - 11x+28=40( phan h = mtinh na) => X= {-1;12}
x2 - 11x+28=-42 (loai)
2x(8x-1)2(4x-1)=9 => (8x-1)2 x (8x2-2x)=9 => (64x2 -16x+1) x (8x2 - 2x)=9
đặt t=(8x2 - 2x) => 64x2 - 16x =8t
ta đc: (8t+1) x t=9 <=> 8t2 + t-9 = 0 => (t-1)(8t+9)=0
=> tu lam na :)
Vậy phương trình có 2nghiệm x=-1/4và x=1/2
phuong trinh dau tien khong xac dinh dc dau tao tra tren mang k co
(x+2)(x+3)(x-5)(x-6)=1802)
=> (x+2)(x-5).(x+3)(x-6) = 180
=> (x²-3x-10)(x²-3x-18) = 180
đặt t = x² - 3x ; ta có pt: (t -10)(t -18) = 180
=> t² - 28t + 180 = 180
=> t² - 28t = 0 => t(t-28)=0 =>t = 0 hoặc t = 28
vi t = x²-3x=> co * x²-3x=0 => x = 0 hoặc x = 3
* x²-3x = 28 <=> x²-3x-28 = 0 <=> x = -4 hoặc x = 7
Vậy pt có 4 nghiệm là: x = 0; x = 3 ; x = -4 ; x = 7