ta có : x^5+2x^4+3x^3+3x^2+2x+1=0

\(\Leftrightarrow\)x^5+x^4+x^4+x^3+2x^3+2x^2+x^2+x+x+1=0

\(\Leftrightarrow\)(x^5+x^4)+(x^4+x^3)+(2x^3+2x^2)+(x^2+x)+(x+1)=0

\(\Leftrightarrow\)x^4(x+1)+x^3(x+1)+2x^2(x+1)+x(x+1)+(x+1)=0

\(\Leftrightarrow\)(x+1)(x^4+x^3+2x^2+x+1)=0

\(\Leftrightarrow\)(x+1)(x^4+x^3+x^2+x^2+x+1)=0

\(\Leftrightarrow\)(x+1)[x^2(x^2+x+1)+(x^2+x+1)]=0

\(\Leftrightarrow\)(x+1)(x^2+x+1)(x^2+1)=0

VÌ x^2+x+1=(x+\(\dfrac{1}{2}\))^2+\(\dfrac{3}{4}\)\(\ne0\) và x^2+1\(\ne0\)

\(\Rightarrow\)x+1=0

\(\Rightarrow\)x=-1

CÒN CÂU B TỰ LÀM (02042006)

b: x^4+3x^3-2x^2+x-3=0

=>x^4-x^3+4x^3-4x^2+2x^2-2x+3x-3=0

=>(x-1)(x^3+4x^2+2x+3)=0

=>x-1=0

=>x=1

Ta có \(\frac{12}{x^2+2x+4}-\frac{5}{x^2+2x+5}=2\)

<=>\(12\left(x^2+2x+5\right)-5\left(x^2+2x+4\right)=2\left(x^2+2x+5\right)\left(x^2+2x+4\right)\)

\(\Leftrightarrow12x^2+24x+60-5x^2-10x-20=2x^4+8x^3+26x^2+36x+40\)

\(\Leftrightarrow7x^2+14x+40=2x^4+8x^3+26x^2+36x+40\)

\(\Leftrightarrow2x^4+8x^3+19x^2+22x=0\)

\(\Leftrightarrow x\left(2x^3+8x^2+19x+22\right)=0\)

\(\Leftrightarrow x\left(2x^3+4x^2+4x^2+8x+11x+22\right)=0\)

\(\Leftrightarrow x\left[2x^2\left(x+2\right)+4x\left(x+2\right)+11\left(x+2\right)\right]=0\)

\(\Leftrightarrow x\left(x+2\right)\left(2x^2+4x+11\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-2\end{cases}}}\)

Vậy PT có nghiệm duy nhất S ={0 ; -2 }  vì(  \(2x^2+4x+11\ne0\))

Đặt 3-x = a ; 2-x = b

=> 5-2x = a+b

pt <=> a^4+b^4 = (a+b)^4 = a^4+4a^3b+6a^2b^2+4ab^3+b^4

<=> a^4+4a^3b+6a^2b^2+4ab^3+b^4-a^4-b^4 = 0

<=> 4a^3b+6a^2b^2+4ab^3 = 0

<=> 2a^3b+3a^2b^2+2ab^3 = 0

<=> ab.(2a^2+3ab+2b^2) = 0

<=> ab=0 ( vì 2a^2+3ab+2b^2 > 0 )

<=> a=0 hoặc b=0

<=> 3-x=0 hoặc 2-x=0

<=> x=3 hoặc x=2

Vậy .............

Tk mk nha

\(\left(2x+4\right)\left(x-3\right)-\left(x+2\right)\left(x-4\right)=x\left(x+5\right)\)

\(2\left(x+2\right)\left(x-3\right)-\left(x+2\right)\left(x-4\right)=x\left(x+5\right)\)

\(\left(x+2\right)\left(2x-6-x+4\right)=x\left(x+5\right)\)

\(\left(x+2\right)\left(x-2\right)-x^2-5x=0\)

\(x^2-2x+2x-4-x^2-5x=0\)

\(-5x-4=0\)

\(-5x=4\)

\(\Rightarrow\)\(x=\frac{-4}{5}\)

\(\left(x-2\right)^2=\left(2x-4\right)\left(x+5\right)\)

\(\left(x-2\right)^2-2\left(x-2\right)\left(x+5\right)=0\)

\(\left(x-2\right)\left(x-2-2x-10\right)=0\)

\(\left(x-2\right)\left(-x-12\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x-2=0\\-x-12=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-12\end{cases}}}\)

Bạn tự kết luận 2 câu nhé

làm hệ PT nghĩ ngay đến xét hàm

ĐKXĐ: \(x\notin\left\{-3;1\right\}\)

Ta có: \(\frac{4}{x^2+2x-3}=\frac{2x-5}{x+3}-\frac{2x}{x-1}\)

\(\Leftrightarrow\frac{\left(2x-5\right)\left(x-1\right)}{\left(x+3\right)\left(x-1\right)}-\frac{2x\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}=\frac{4}{\left(x-1\right)\left(x+3\right)}\)

Suy ra: \(\left(2x-5\right)\left(x-1\right)-2x\left(x+3\right)=4\)

\(\Leftrightarrow2x^2-2x-5x+5-2x^2-6x=4\)

\(\Leftrightarrow-13x+5=4\)

\(\Leftrightarrow-13x=4-5=-1\)

hay \(x=\frac{1}{13}\)(nhận)

Vậy: \(S=\left\{\frac{1}{13}\right\}\)

giải phaj bỏ ngoặc nhức đầu lắm

\(\hept{\begin{cases}\left|2x+5\right|=2x+5\Leftrightarrow x\ge-\frac{5}{2}\\\left|2x+5\right|=-\left(2x+5\right)\Leftrightarrow x< -\frac{5}{2}\end{cases}}\)

\(\hept{\begin{cases}\left|4-x\right|=4-x\Leftrightarrow x\le4\\\left|4-x\right|=x-4\Leftrightarrow x>4\end{cases}}\)

\(\hept{\begin{cases}\left|x+9\right|=x+9\Leftrightarrow x\ge-9\\\left|x+9\right|=-\left(x+9\right)\Leftrightarrow x< -9\end{cases}}\)

(+) \(-\frac{5}{2}\le x\le4\) \(\left(-\frac{5}{2}>-9\right)\)

\(pt\Leftrightarrow2x+5+4-x=x+9\)

\(\Leftrightarrow0x=0\left(true\right)\)

(+) \(-9\le x< -\frac{5}{2}\) \(\left(-\frac{5}{2}< 4\right)\)

\(pt\Leftrightarrow-\left(2x+5\right)+4-x=x+9\)

\(\Leftrightarrow-2x-5+4-x=x+9\)

\(\Leftrightarrow-4x=10\Leftrightarrow x=-\frac{5}{2}\)( không thỏa mãn )

Vậy phương trình nhận mọi x trong khoảng \(-\frac{5}{2}\le x\le4\)làm nghiệm

Ta có |2x + 5| + |4 - x| = |x + 9|

=> \(\orbr{\begin{cases}\left|2x+5\right|+\left|4-x\right|=x+9\\\left|2x+5\right|+\left|4-x\right|=-x-9\end{cases}}\)

Khi |2x + 5| + |4 - x| = x + 9 (1)

Nếu x < -2,5

=> |2x + 5| = - (2x + 5) = -2x - 5

=> |4 - x| = 4 - x

=> (1) <=> -2x - 5 + 4 - x = x + 9

=> -2x - x - x = 9 - 4 + 5

=> - 4x = 10

=> x = -2,5 (loại)

Nếu \(-2,5\le x\le4\)

=> |2x + 5| = 2x + 5

|4 - x| = 4 - x

=> (1) <=> 2x + 5 + 4 - x = x + 9

=> 2x - x - x = 9 - 5 - 4

=> 0x = 0

=> x thỏa mãn với  \(-2,5\le x\le4\)

Nếu x > 4

=> |2x + 5| = 2x + 5

|4 - x| = -4 + x

=> (1) <=> 2x + 5 - 4 + x = x + 9

=> 2x + x - x = 9 - 5 + 4

=> 2x = 8

=> x = 4 (loại)

Vậy khi |2x + 5| + |4 - x| = x + 9 thì  \(-2,5\le x\le4\)

Khi |2x + 5| + |4 - x| = -x - 9

Nếu x < -2,5

=> |2x + 5| = - (2x + 5) = -2x - 5

=> |4 - x| = 4 - x

=> (1) <=> -2x - 5 + 4 - x = -x - 9

=> -2x - x + x = -9 - 4 + 5

=> - 2x = -8

=> x = 4 (loại)

Nếu \(-2,5\le x\le4\)

=> |2x + 5| = 2x + 5

|4 - x| = 4 - x

=> (1) <=> 2x + 5 + 4 - x = -x - 9

=> 2x - x + x = -9 - 5 - 4

=> 2x = -18

=> x = -9 (loại)

Nếu x > 4

=> |2x + 5| = 2x + 5

|4 - x| = -4 + x

=> (1) <=> 2x + 5 - 4 + x = - x - 9

=> 2x + x + x = 9 - 5 + 4

=> 4x = 8

=> x = 2 (loại)

Vậy khi |2x + 5| + |4 - x| = -x - 9 thì \(x\in\varnothing\)

Vậy  \(-2,5\le x\le4\)

TH1:

\(x>0\Leftrightarrow\hept{\begin{cases}2x+5>5\\4-x>4\\x+9>9\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}|2x-5|=2x-5\\|4-x|=4-x\\|x=9|=x+9\end{cases}}\)

\(\Leftrightarrow2x+5+4-x=x+9\)

\(\Leftrightarrow2x-x-x=9-4-5\)

\(\Leftrightarrow0x=0\)(thỏa mãn điều kiện);

vậy pt vô số nghiệm

TH2:

\(x< 0\Leftrightarrow\hept{\begin{cases}2x+5< 5\\4-x< 4\\x+9< 9\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}|2x+5|=-\left(2x+5\right)=-2x-5\\|4-x|=-\left(4-x\right)=x-4\\|x+9|=-\left(x+9\right)=-x-9\end{cases}}\)

\(\Leftrightarrow-2x-5+x-4=-x-9\)

\(\Leftrightarrow-2x+x+x=4+5-9\)

\(\Leftrightarrow0x=0\)(thỏa mãn điều kiện);

vậy pt vô số nghiệm