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\(\dfrac{x+1}{x^2+x+1}-\dfrac{x-1}{x^2-x+1}=\dfrac{3}{x\left(x^4+x^2+1\right)}\)
\(\Leftrightarrow\dfrac{x\left(x+1\right)\left(x^2-x+1\right)-x\left(x-1\right)\left(x^2+x+1\right)}{x\left(x^2+x+1\right)\left(x^2-x+1\right)}=\dfrac{3}{x\left(x^2+x+1\right)\left(x^2-x+1\right)}\)
\(\Leftrightarrow x\left(x^3+1\right)-x\left(x^3-1\right)=3\)
=>2x=3
hay x=3/2
\(\left(x-1\right)^2-1+x^2=\left(1-x\right).\left(x+3\right)\)
\(x^2-2x+1-1+x^2=x-x^2+3-3x\)
\(2x^2-2x=-2x-x^2+3\)
\(2x^2+x^2-2x+2x=3\)
\(3x^2=3\)
\(\Rightarrow x^2=1\Rightarrow x=\left\{1;-1\right\}\)
(x-1)^2 - 1 + x^2= (1-x)(x+3)
\(\Leftrightarrow\)x^2-2x+1-1+x^2=x+3-x^2+3x
\(\Leftrightarrow\)2x^2-2x=4x-x^2+3
\(\Leftrightarrow\)3x^2-6x=3
\(\Leftrightarrow\)3x(x-2)=3
\(\orbr{\begin{cases}3x=3\\x-2=3\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=5\end{cases}}\)
Vậy pt có 2 nghiệm x=1 và x=5
VT=(x-1)3+(2-x)(4+2x+x2)+3x(x+2)=9x+7 (*)
thay (*) vào VT của pt đầu ta đc
=>9x+7=17
=>9x=10
=>x=\(\frac{10}{9}\)
\(a,\left(x-2\right)^2-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=15\)\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6\left(x^2+2x+1\right)=15\)\(\Leftrightarrow-6x^2+12x+19+6x^2+12x+6=15\)
\(\Leftrightarrow24x=-10\)
\(\Leftrightarrow x=-\dfrac{5}{12}\)
Vậy:....
\(b,\left(5x+1\right)^2-\left(5x+3\right)\left(5x-3\right)=30\)
\(\Leftrightarrow25x^2+10x+1-25^2+9=30\)
\(\Leftrightarrow10x=20\)
\(\Rightarrow x=2\)
Vậy :....
\(c,\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-2\right)\left(x+2\right)=15\)\(\Leftrightarrow x^3+27-x\left(x^2-4\right)=15\)
\(\Leftrightarrow x^3+27-x^3+4x=15\)
\(\Leftrightarrow4x=15-27=-12\)
\(\Leftrightarrow x=-3\)
vậy : .....
Lời giải:
\((x^3-x^2)-4x^2+8x-4=0\)
\(\Leftrightarrow x^2(x-1)-4(x^2-2x+1)=0\)
\(\Leftrightarrow x^2(x-1)-4(x-1)^2=0\)
\(\Leftrightarrow (x-1)[x^2-4(x-1)]=0\)
\(\Leftrightarrow (x-1)(x^2-4x+4)=0\)
\(\Leftrightarrow (x-1)(x-2)^2=0\)
\(\Rightarrow \left[\begin{matrix} x=1\\ x=2\end{matrix}\right.\)
- \(\left(2x+5\right)^2=\left(x+2\right)^2\Leftrightarrow\left(2x+5+x+2\right)\left(2x+5-x-2\right)=0\)\(\Leftrightarrow\left(3x+7\right)\left(x+3\right)=0\Leftrightarrow\orbr{\begin{cases}x=-\frac{7}{3}\\x=-3\end{cases}}\)
- \(x^2-5x+6=0\Leftrightarrow x^2-6x+x-6=0\Leftrightarrow x\left(x-6\right)+\left(x-6\right)=0\)\(\left(x+1\right)\left(x-6\right)=0\Leftrightarrow\orbr{\begin{cases}x=6\\x=-1\end{cases}}\)
- \(2x^3+6x^2=x^2+3x\Leftrightarrow2x^2\left(x+3\right)=x\left(x+3\right)\)\(\Leftrightarrow\left(x+3\right)\left(2x^2-x\right)=0\Leftrightarrow x\left(2x-1\right)\left(x+3\right)=0\Leftrightarrow\)\(x=0\)hoặc \(x=\frac{1}{2}\)hoặc \(x=-3\)
Ap dung BDT Bunhiacopxki , ta co :
( x2 + y2)2 = ( \(\sqrt{x^4}+\sqrt{y^4}\))2 = \(\left(\sqrt{x}.\sqrt{x^3}+\sqrt{y}.\sqrt{y^3}\right)\)2 ≤ ( x+y)( x3 + y3) = 2(x+ y)
⇔ ( x2 + y2)2 ≤ 2( x + y)
⇔ ( x2 + y2)4 ≤ 4( x + y)2 ≤ 4( x2 + y2)( 12 + 12) = 8( x2 + y2)
⇔ ( x2 + y2)4 ≤ 8( x2 + y2)
⇔ ( x2 + y2)3 ≤ 8
⇔ x2 + y2 ≤ 2
Dau " =" xay ra khi : x = y = 1
P/s : Mk lam thu thui nha , khong chac dau
Đời về bản là buồn... cười!!!Phùng Khánh LinhHong Ra Onchú tuổi gìNguyễn Ngô Minh TríNhã Doanh,.....
Mk can gap gap , mai thi hoc ky 2 rui nhen
Đặt a = x2 + 3x - 4 ; b = 2x2 - 5x + 3
=> 3x2 - 2x - 1 = a + b
khi đó phương trình đã cho có dạng: a3 + b3 = (a+ b)3
=> a3 + b3 = a3 + b3 + 3ab(a + b) => 3ab (a+b) = 0 => a= 0 hoặc b = 0 hoặc a = -b
Nếu a = 0 => x2 + 3x - 4 = 0 => x2 + 4x- x - 4 = 0 => (x - 1)(x + 4) = 0 => x = 1; -4
Nếu b = 0 => 2x2 - 5x + 3 = 0 => 2x2 - 2x - 3x + 3 = 0 => (2x-3)(x - 1) = 0 => x = 3/2; 1
Nếu a = - b => - (2x2 - 5x + 3) = x2 + 3x - 4 => 3x2 - 2x - 1 = 0 => 3x2 - 3x + x - 1 = 0 => (3x + 1)(x - 1) = 0 => x = -1/3; 1
Vậy x = 1; 3/2; -1/3; -4
Pt ⇔4x2+x+3+4xx+3−−−−√+2x−1+1−22x−1−−−−−√=0⇔(2x−x+3−−−−√)2−√−1)2=0⇔x=1⇔4x2+x+3+4xx+3+2x−1+1−22x−1=0⇔(2x−x+3)2+(2x−1−1)2=0⇔x=1
a, x2 ( x+4.5)=13.5
x2 (x+4.5)= 56
x2 (x+20)=56
Tự làm típ đi nhé
a) x2(x+4.5)=13.5
x=1,5