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

b)\(\sqrt{25x^2}=19\)

\(\Leftrightarrow5x=19\)

\(\Leftrightarrow x=\dfrac{19}{5}\)

c)\(\sqrt{x-7}+3=0\)

\(\Leftrightarrow\sqrt{x-7}=-3\)

\(\Leftrightarrow x-7=9\)

\(\Leftrightarrow x=16\)

Giải PT

a) \(3\sqrt{9x}+\sqrt{25x}-\sqrt{4x} = 3\)

\(\Leftrightarrow\) \(3.3\sqrt{x} +5\sqrt{x} - 2\sqrt{x} = 3 \)

\(\Leftrightarrow\) \(9\sqrt{x}+5\sqrt{x}-2\sqrt{x} = 3 \)

\(\Leftrightarrow\) \(12\sqrt{x} = 3\)

\(\Leftrightarrow\) \(\sqrt{x} = 4 \)

\(\Leftrightarrow\) \(\sqrt{x^2} = 4^2\)

\(\Leftrightarrow\) \(x=16\)

b) \(\sqrt{x^2-2x-1} - 3 =0\)

\(\Leftrightarrow\) \(\sqrt{(x-1)^2} -3=0\)

\(\Leftrightarrow\) \(|x-1|=3\)

* \(x-1=3\)

\(\Leftrightarrow\) \(x=4\)

* \(-x-1=3\)

\(\Leftrightarrow\) \(-x=4\)

\(\Leftrightarrow\) \(x=-4\)

c) \(\sqrt{4x^2+4x+1} - x = 3\)

<=> \(\sqrt{(2x+1)^2} = 3+x\)

<=> \(|2x+1|=3+x\)

* \(2x+1=3+x\)

<=> \(2x-x=3-1\)

<=> \(x=2\)

* \(-2x+1=3+x\)

<=> \(-2x-x = 3-1\)

<=> \(-3x=2\)

<=> \(x=\dfrac{-2}{3}\)

d) \(\sqrt{x-1} = x-3\)

<=> \(\sqrt{(x-1)^2} = (x-3)^2\)

<=> \(|x-1| = x^2-2.x.3+3^2\)

<=> \(|x-1| = x-6x+9\)

<=> \(|x-1| = -5x+9\)

* \(x-1= -5x+9\)

<=> \(x+5x = 9+1\)

<=> \(6x=10\)

<=> \(x= \dfrac{10}{6} =\dfrac{5}{3}\)

* \(-x-1 = -5x+9\)

<=> \(-x+5x = 9+1\)

<=> \(4x = 10\)

<=> \(x= \dfrac{10}{4} = \dfrac{5}{2}\)

mình nghĩ câu b \(\left(x-1\right)^2\)luôn lớn hơn 0 nên chắc không cần chia ra hai trường hợp nhỉ ?

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

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

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