làm tạm câu này vậy

a/\(\left(x^2-x+1\right)^4+4x^2\left(x^2-x+1\right)^2=5x^4\)

\(\Leftrightarrow\left(x^2-x+1\right)^4+4x^2\left(x^2-x+1\right)+4x^4=9x^4\)

\(\Leftrightarrow\left\{\left(x^2-x+1\right)^2+2x^2\right\}=\left(3x^2\right)^2\)

\(\Leftrightarrow\left(x^2-x+1\right)^2+2x^2=3x^2\)(vì 2 vế đều không âm)

\(\Leftrightarrow\left(x^2-x+1\right)=x^2\)

\(\Leftrightarrow\left|x\right|=x^2-x+1\)\(\left(x^2-x+1=\left(x-\frac{1}{4}\right)^2+\frac{3}{4}>0\right)\)

\(\Leftrightarrow\orbr{\begin{cases}x=x^2-x+1\\-x=x^2-x+1\end{cases}\Leftrightarrow\orbr{\begin{cases}\left(x-1\right)^2=0\\x^2+1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\\x^2+1=0\left(vo.nghiem\right)\end{cases}}}\)

Vậy...

chuẩn

a: =>(x^2+4x-5)(x^2+4x-21)=297

=>(x^2+4x)^2-26(x^2+4x)+105-297=0

=>x^2+4x=32 hoặc x^2+4x=-6(loại)

=>x^2+4x-32=0

=>(x+8)(x-4)=0

=>x=4 hoặc x=-8

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

hay \(x\in\left\{\dfrac{1+\sqrt{13}}{2};\dfrac{1-\sqrt{13}}{2};\dfrac{-1+\sqrt{17}}{2};\dfrac{-1-\sqrt{17}}{2}\right\}\)

c: =>(x-1)(x+2)(x^2-6x-2)=0

hay \(x\in\left\{1;-2;3+\sqrt{11};3-\sqrt{11}\right\}\)

Bài 1:

a) \(\Delta=b^2-4ac=\left(-5\right)^2-4\cdot2\cdot1=25-8=17\)

Vì Δ>0 nên phương trình \(2x^2-5x+1=0\) có hai nghiệm là:

\(\left\{{}\begin{matrix}x_1=\frac{-b-\sqrt{\Delta}}{2a}\\x_2=\frac{-b+\sqrt{\Delta}}{2a}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_1=\frac{5-\sqrt{17}}{2\cdot2}=\frac{5-\sqrt{17}}{4}\\x_2=\frac{5+\sqrt{17}}{2\cdot2}=\frac{5+\sqrt{17}}{4}\end{matrix}\right.\)

Vậy: \(S=\left\{\frac{5-\sqrt{17}}{4};\frac{5+\sqrt{17}}{4}\right\}\)

b) Ta có: \(4x^2+4x+1=0\)

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

\(\Leftrightarrow2x+1=0\)

\(\Leftrightarrow2x=-1\)

hay \(x=-\frac{1}{2}\)

Vậy: \(S=\left\{\frac{-1}{2}\right\}\)

c) Ta có: \(-3x^2+2x+8=0\)

\(\Leftrightarrow-3x^2+6x-4x+8=0\)

\(\Leftrightarrow-3x\left(x-2\right)-4\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(-3x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\-3x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\-3x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{-4}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{2;\frac{-4}{3}\right\}\)

d) Ta có: \(5x^2-6x-1=0\)

\(\Delta=b^2-4\cdot a\cdot c=\left(-6\right)^2-4\cdot5\cdot\left(-1\right)=56\)

Vì Δ>0 nên phương trình \(5x^2-6x-1=0\) có hai nghiệm là:

\(\left\{{}\begin{matrix}x_1=\frac{-b-\sqrt{\Delta}}{2a}\\x_2=\frac{-b+\sqrt{\Delta}}{2a}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_1=\frac{6-\sqrt{56}}{2\cdot5}=\frac{3-\sqrt{14}}{5}\\x_2=\frac{6+\sqrt{56}}{2\cdot5}=\frac{3+\sqrt{14}}{5}\end{matrix}\right.\)

Vậy: \(S=\left\{\frac{3-\sqrt{14}}{5};\frac{3+\sqrt{14}}{5}\right\}\)

e) Ta có: \(-3x^2+14x-8=0\)

\(\Leftrightarrow-3x^2+12x+2x-8=0\)

\(\Leftrightarrow-3x\left(x-4\right)+2\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(-3x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\-3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\-3x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\frac{2}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{4;\frac{2}{3}\right\}\)

g) Ta có: \(-7x^2+4x-3=0\)

\(\Delta=b^2-4ac=4^2-4\cdot\left(-7\right)\cdot\left(-3\right)=-68\)

Vì Δ<0 nên phương trình \(-7x^2+4x-3=0\) không có nghiệm

Vậy: S=∅

Cảm ơn nhá

a, \(\Delta=25-8=17\)>0 Vậy pt có 2 nghiệm pb 

\(x=\dfrac{5\pm\sqrt{17}}{4}\)

b, \(\Delta=16-16=0\)Vậy pt có nghiệm kép 

\(x_1=x_2=\dfrac{1}{4}\)

c, \(\Delta=1-4.2.5< 0\)Vậy pt vô nghiệm 

d, \(\Delta=4+4.24=100>0\)Vậy pt có 2 nghiệm pb 

\(x=\dfrac{-2-10}{-6}=2;x=\dfrac{-2+10}{-6}=-\dfrac{4}{3}\)

cảm ơn bạn

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

\(\left(x^5-x^4\right)-\left(4x^4-4x^3\right)+\left(4x^2-4x\right)-\left(x-1\right)=0\)

\(x^4\left(x-1\right)-4x^3\left(x-1\right)+4x\left(x-1\right)-\left(x-1\right)=0\)

\(\left(x-1\right)\left(x^4-4x^3+4x-1\right)=0\)

\(\left(x-1\right)\left[\left(x^4-1\right)-\left(4x^3-4x\right)\right]=0\)

\(\left(x-1\right)\left[\left(x-1\right)\left(x^3+x^2+x+1\right)-4x\left(x^2-1\right)\right]=0\)

\(\left(x-1\right)\left[\left(x-1\right)\left(x^3+x^2+x+1\right)-4x\left(x-1\right)\left(x+1\right)\right]=0\)

\(\left(x-1\right)^2\left(x^3+x^2+x+1-4x^2-4x\right)=0\)

\(\left(x-1\right)^2\left(x^3-3x^2-3x+1\right)=0\)

\(\left(x-1\right)^2\left[\left(x+1\right)\left(x^2-x+1\right)-3x\left(x+1\right)\right]=0\)

\(\left(x-1\right)^2\left(x+1\right)\left(x^2-x+1-3x\right)=0\)

\(\left(x-1\right)^2\left(x+1\right)\left[\left(x^2-2.x.2+2^2\right)-3\right]=0\)

\(\left(x-1\right)^2\left(x+1\right)\left[\left(x-2\right)^2-\left(\sqrt{3}\right)^2\right]=0\)

\(\left(x-1\right)^2\left(x+1\right)\left(x-2-\sqrt{3}\right)\left(x-2+\sqrt{3}\right)=0\)

Đến đây b tự làm tiếp nhé~

1) -x²+4x-6+ \(\frac{21}{x^2-4x+10}\)= 0

Đặt -x²+4x+10 là a, ta có:

-a +4+\(\frac{21}{a}\)=0

=> \(\frac{21+4a-a^2}{a}\)=0

=> 21+4a-a²=0

=>-(a-2)²=-25

=> (a-2)²=25 => \(\orbr{\begin{cases}a=7\\a=-3\end{cases}}\)

Bạn thay a vào rồi tính tiếp nha

a: ĐKXĐ: x∉{5;-5}

Ta có: \(\frac{2}{x-5}+\frac{3}{x+5}+\frac{-2x+20}{x^2-25}=0\)

=>\(\frac{2}{x-5}+\frac{3}{x+5}+\frac{-2x+20}{\left(x-5\right)\left(x+5\right)}=0\)

=>\(\frac{2\left(x+5\right)+3\left(x-5\right)-2x+20}{\left(x-5\right)\left(x+5\right)}=0\)

=>2(x+5)+3(x-5)-2x+20=0

=>2x+10+3x-15-2x+20=0

=>3x+15=0

=>3x=-15

=>x=-5(loại)

b: ĐKXĐ: x∉{2;-2}

Ta có: \(\frac{3x}{x-2}+\frac{4x}{x+2}+\frac{-5x^2-2x}{x^2-4}=0\)

=>\(\frac{3x}{x-2}+\frac{4x}{x+2}+\frac{-5x^2-2x}{\left(x-2\right)\left(x+2\right)}=0\)

=>\(\frac{3x\left(x+2\right)+4x\left(x-2\right)-5x^2-2x}{\left(x-2\right)\left(x+2\right)}=0\)

=>\(3x\left(x+2\right)+4x\left(x-2\right)-5x^2-2x=0\)

=>\(3x^2+6x+4x^2-8x-5x^2-2x=0\)

=>\(2x^2-4x=0\)

=>2x(x-2)=0

=>x(x-2)=0

=>\(\left[\begin{array}{l}x=0\\ x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\left(nhận\right)\\ x=2\left(loại\right)\end{array}\right.\)

Tưởng bn lớp 5 ạ?? Sao lại đăng câu hỏi lp 9 ạ??:)

minh lop 5 dang chi minh muon nick cua minh