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1. \(x^2\left(x+1\right)+x+1=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow x+1=0\Rightarrow x=-1\)
2. \(\left(x-2\right)\left(6x+2\right)+\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(x-2\right)\left(6x+2+x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right).7x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\7x=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\)
3.
\(x^2-5x+6=0\)
\(\Leftrightarrow x^2-2x-3x+6=0\)
\(\Leftrightarrow x\left(x-2\right)-3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
4.
\(x^2-x-6=0\)
\(\Leftrightarrow x^2+2x-3x-6=0\)
\(\Leftrightarrow x\left(x+2\right)-3\left(x+2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
a, \(\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=16\)
\(\Leftrightarrow x^2+8x+16-\left(x^2-x+x-1\right)=16\)
\(\Leftrightarrow8x+1=0\Leftrightarrow x=-\frac{1}{8}\)
b, \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)
\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5\left(x^2-49\right)=0\)
\(\Leftrightarrow2x+255=0\Leftrightarrow x=-\frac{225}{2}\)
c, \(\left(x+2\right)\left(x-2\right)-x^3-2x=15\)
\(\Leftrightarrow x^2-4-x^3-2x=15\)( vô nghiệm )
d, \(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)=28\)
\(\Leftrightarrow x^3+9x^2+27x+27-9x^3+6x^2-x+8x^3+1=28\)
\(\Leftrightarrow15x^2+26=0\Leftrightarrow x^2\ne-\frac{26}{15}\)( vô nghiệm )
Tính nhẩm hết á, sai bỏ quá nhá, sắp đi hc ... nên chất lượng hơi kém xíu ~~~
1,(x+2)(x+5)(x+3)(x+4)-24=(x2+7x+10)(x2+7x+12)-24
Đặt x2+7x+10= t ta có t(t+2)-24=t2+2t-24=(t-4)(t+6)
hay (x2+7x+6)(x2+7x+16)
2,x(x+10)(x+4)(x+6)+128=(x2+10x)(x2+10x+24)+128
Đặt x2+10x=t ta có t(t+24)+128=t2+24t+128=(t+8)(t+16)
hay (x2+10x+8)(x2+10x+16)
3,(x+2)(x-7)(x+3)(x-8)-144=(x2-5x-14)(x2-5x-24)-144
Đặt x2-5x-14=t ta có t(t-10)-144=t2-10t-144=(t-18)(t+8)
Hay (x2-5x-32)(x2-5x-6)=(x2-5x-32)(x+1)(x-6)
\(b,\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\) \(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5x^2+245=0\)\(\Leftrightarrow2x=-255\Rightarrow x=-\dfrac{255}{2}\)
\(c,\left(x-2\right)^3-\left(x-4\right)\left(x^2+4x+16\right)+6\left(x+1\right)^2=49\)\(\Leftrightarrow x^3-6x^2+12x-8-x^3+64+6x^2+12x+6-49=0\)\(\Leftrightarrow24x=-13\Rightarrow x=-\dfrac{13}{24}\)
\(d,\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)
\(\Leftrightarrow x^3+8-x^3-2x=15\)
\(\Leftrightarrow-2x=23\Rightarrow x=-\dfrac{23}{2}\)
a) \(\left(x+1\right)\left(2x-1\right)\left(-x+2\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}x+1=0\\2x-1=0\\-x+2=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=-1\\x=\frac{1}{2}\\x=2\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là \(S=\left\{-1;\frac{1}{2};2\right\}\)
b) \(\left(2x-1\right)\left(3x+2\right)\left(4x-5\right)\left(x-7\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}2x-1=0\\3x+2=0\\4x-5=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=\frac{1}{2}\\x=-\frac{2}{3}\\x=\frac{5}{4}\\x=7\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là \(S=\left\{\frac{1}{2};-\frac{2}{3};\frac{5}{4};7\right\}\)
c) \(x^2-6x+11=0\)
\(\Leftrightarrow x^2-6x+9+2=0\)
\(\Leftrightarrow\left(x-3\right)^2+2=0\) (vô lí)
Vậy phương trình vô nghiệm
d) \(\left(x^2+2x+3\right)\left(x^2-25\right)\left(x+19\right)=0\)
\(\Leftrightarrow\left(x^2+2x+1+2\right)\left(x+5\right)\left(x-5\right)\left(x+19\right)=0\)
\(\Leftrightarrow\left[\left(x+1\right)^2+2\right]\left(x+5\right)\left(x-5\right)\left(x+19\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}x+5=0\\x-5=0\\x+19=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=-5\\x=5\\x=-19\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là \(S=\left\{\pm5;-19\right\}\)
a,b,d dễ mà bạn tự làm
c,x2-6x+11=0<=> x2-6x+9+2=0
<=>(x-3)2=-2(vô lý)
vậy pt vô nghiệm
a: \(\Leftrightarrow x^3-3x^2+3x-1-x^3+2x^2-x=5x\left(2-x\right)-11\left(x+2\right)\)
=>-x^2+2x-1=10x-5x^2-11x-22
=>-x^2+2x-1=-5x^2-x-22
=>4x^2+3x+21=0
=>PTVN
b: \(\Leftrightarrow\left(x+10\right)\left(x+4\right)+3\left(x+4\right)\left(x-2\right)=4\left(x+10\right)\left(x-2\right)\)
=>x^2+14x+40+3(x^2+2x-8)=4(x^2+8x-20)
=>x^2+14x+40+3x^2+6x-24=4x^2+32x-80
=>20x+16=32x-80
=>-12x=-96
=>x=8
c: \(\Leftrightarrow6\left(x-3\right)+7\left(x-5\right)=13x+4\)
=>6x-18+7x-35=13x+4
=>-53=4(loại)
d: =>3(2x-1)-5(x-2)=3(x+7)
=>6x-3-5x+10=3x+21
=>3x+21=x+7
=>x=-7
e: =>x^3-6x^2+12x-8-x^3-3x^2-3x-1=-9x^2+1
=>-9x^2+9x-9=-9x^2+1
=>9x=10
=>x=10/9
hỏi j thế bạn, không có đề sao làm
đề bài sai