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\(1.\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}.\Leftrightarrow\dfrac{x-1-3x}{3}=\dfrac{x-2}{2}.\Leftrightarrow\dfrac{-2x-1}{3}-\dfrac{x-2}{2}=0.\)
\(\Leftrightarrow\dfrac{-4x-2-3x+6}{6}=0.\Rightarrow-7x+4=0.\Leftrightarrow x=\dfrac{4}{7}.\)
\(2.\left(x-2\right)\left(2x-1\right)=x^2-2x.\Leftrightarrow\left(x-2\right)\left(2x-1\right)-x\left(x-2\right)=0.\)
\(\Leftrightarrow\left(x-2\right)\left(2x-1-x\right)=0.\Leftrightarrow\left(x-2\right)\left(x-1\right)=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2.\\x=1.\end{matrix}\right.\)
\(3.3x^2-4x+1=0.\Leftrightarrow\left(x-1\right)\left(x-\dfrac{1}{3}\right)=0.\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=\dfrac{1}{3}.\end{matrix}\right.\)
\(4.\left|2x-4\right|=0.\Leftrightarrow2x-4=0.\Leftrightarrow x=2.\)
\(5.\left|3x+2\right|=4.\Leftrightarrow\left[{}\begin{matrix}3x+2=4.\\3x+2=-4.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}.\\x=-2.\end{matrix}\right.\)
\(1,\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}\\ \Leftrightarrow\dfrac{x-1}{3}-x=\dfrac{x-2}{2}\\ \Leftrightarrow\dfrac{2\left(x-1\right)-6x}{6}=\dfrac{3\left(x-2\right)}{6}\\ \Leftrightarrow2\left(x-1\right)-6x=3\left(x-2\right)\\ \Leftrightarrow2x-2-6x=3x-6\\ \Leftrightarrow-4x-2=3x-6\)
\(\Leftrightarrow3x-6+4x+2=0\\ \Leftrightarrow7x-4=0\\ \Leftrightarrow x=\dfrac{4}{7}\)
\(2,\left(x-2\right)\left(2x-1\right)=x^2-2x\\ \Leftrightarrow2x^2-4x-x+2=x^2-2x\\ \Leftrightarrow x^2-3x+2=0\\ \Leftrightarrow\left(x^2-2x\right)-\left(x-2\right)=0\\ \Leftrightarrow x\left(x-2\right)-\left(x-2\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
\(3,3x^2-4x+1=0\\ \Leftrightarrow\left(3x^2-3x\right)-\left(x-1\right)=0\\ \Leftrightarrow3x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(3x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)
\(4,\left|2x-4\right|=0\\ \Leftrightarrow2x-4=0\\ \Leftrightarrow2x=4\\ \Leftrightarrow x=2\)
\(5,\left|3x+2\right|=4\\ \Leftrightarrow\left[{}\begin{matrix}3x+2=4\\3x+2=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=2\\3x=-6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-2\end{matrix}\right.\)
\(6,\left|2x-5\right|=\left|-x+2\right|\\ \Leftrightarrow\left[{}\begin{matrix}2x-5=-x+2\\2x-5=x-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=7\\x=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=3\end{matrix}\right.\)
\(\left(3x-4\right)^2-4\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(3x-4\right)^2-\left(2x+2\right)^2=0\)
\(\Leftrightarrow\left(3x-4-2x-2\right)\left(3x-4+2x+2\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(5x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=6\\x=\frac{2}{5}\end{cases}}\) ( thỏa mãn )
Vậy : ...
1/ \(\left(3x-4\right)^2-4\left(x+1\right)^2=0\)
\(\Leftrightarrow9x^2-24x+16-4\left(x^2+2x+1\right)=0\)
\(\Leftrightarrow9x^2-24x+16-4x^2-8x-4=0\)
\(\Leftrightarrow5x^2-32x+12=0\)
\(\Leftrightarrow5x^2-30x-2x+12=0\)
\(\Leftrightarrow5x\left(x-6\right)-2\left(x-6\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(5x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-6=0\\5x-2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=6\\x=\frac{2}{5}\end{cases}}\)
Vậy tập nghiệm của phương trình là : \(S=\left\{6;\frac{2}{5}\right\}\)
2/ \(x^4+2x^3-3x^2-8x-4=0\)
\(\Leftrightarrow x^4+2x^3-3x^2-6x-2x-4=0\)
\(\Leftrightarrow x^3\left(x+2\right)-3x\left(x+2\right)-2\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^3-3x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^3+2x^2+x-2x^2-4x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left[x\left(x^2+2x+1\right)-2\left(x^2+2x+1\right)\right]=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+1\right)^2\left(x-2\right)=0\)
\(\Leftrightarrow\)\(x+2=0\)
hoặc \(x+1=0\)
hoặc \(x-2=0\)
\(\Leftrightarrow\)\(x=2\)
hoặc \(x=-1\)
hoặc \(x=2\)
Vậy tập nghiệm của phương trình là \(S=\left\{2;-2;-1\right\}\)
Với \(x=0\) không phải nghiệm
Với \(x\ne0\) chia 2 vế cho \(x^2\), pt tương đương:
\(2x^2+3x-1+\dfrac{3}{x}+\dfrac{2}{x^2}=0\)
\(\Leftrightarrow2\left(x+\dfrac{1}{x}\right)^2+3\left(x+\dfrac{1}{x}\right)-5=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{x}=1\\x+\dfrac{1}{x}=-\dfrac{5}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x+1=0\\2x^2+5x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\left(vô-nghiệm\right)\\\left(x+2\right)\left(2x+1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-\dfrac{1}{2}\end{matrix}\right.\)
Câu a chắc là đề sai, vì nghiệm vô cùng xấu, tử số của phân thức cuối cùng là \(x+17\) mới hợp lý
b.
Đặt \(x+3=t\)
\(\Rightarrow\left(t+1\right)^4+\left(t-1\right)^4=14\)
\(\Leftrightarrow t^4+6t^2-6=0\) (đến đây đoán rằng bạn tiếp tục ghi sai đề, nhưng thôi cứ giải tiếp)
\(\Rightarrow\left[{}\begin{matrix}t^2=-3+\sqrt{15}\\t^2=-3-\sqrt{15}\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow t=\pm\sqrt{-3+\sqrt{15}}\Rightarrow x=-3\pm\sqrt{-3+\sqrt{15}}\)
Câu c chắc cũng sai đề, vì lên lớp 8 rồi không ai cho đề kiểu này cả, người ta sẽ rút gọn luôn số 1 bên trái và 60 bên phải.
f. 5 – (x – 6) = 4(3 – 2x)
<=>5-x+6=12-8x
<=>7x=1
<=>x=\(\dfrac{1}{7}\)
g. 7 – (2x + 4) = – (x + 4)
<=>7-2x-4=-x-4
<=>x=7
h. 2x(x+2)\(^2\)−8x\(^2\)=2(x−2)(x\(^2\)+2x+4)
<=>\(2x\left(x^2+4x+4\right)-8x^2=2\left(x^3-8\right)\)
<=>\(2x^3+8x^2+8x-8x^2=2\left(x^3-8\right)\)
<=>\(2x^3+8x=2x^3-16\)
<=>\(8x=-16\)
<=>\(x=-2\)
i. (x−2\(^3\))+(3x−1)(3x+1)=(x+1)\(^3\)
<=>\(x-8+9x^2-1=x^3+3x^2+3x+1\)
<=>\(6x^2-2x-10=0\)
<=>\(3x^2-x-5=0\)
<=>\(\left[{}\begin{matrix}x=\dfrac{1+\sqrt{61}}{6}\\x=\dfrac{1-\sqrt{61}}{6}\end{matrix}\right.\)
k. (x + 1)(2x – 3) = (2x – 1)(x + 5)
<=>\(2x^2-x-3=2x^2+9x-5\)
<=>10x=2
<=>\(x=\dfrac{1}{5}\)
f. 5 – (x – 6) = 4(3 – 2x)
<=>5-x+6=12-8x
<=>7x=1
<=>x=\(\dfrac{1}{7}\)
g. 7 – (2x + 4) = – (x + 4)
<=>7-2x-4=-x-4
<=>x=7
h. \(2x\left(x+2\right)^2-8x^2=2\left(x-2\right)\left(x^2+2x+4\right)\)
<=>\(2x\left(x^2+4x+4\right)-8x^2=2\left(x^3-8\right)\)
<=>\(2x^3+8x^2+8x-8x^2=2x^3-16\)
<=>\(8x=-16\)
<=>x=-2
i.\(\left(x-2\right)^3+\left(3x-1\right)\left(3x+1\right)=\left(x+1\right)^3\)
<=>\(x^3-6x^2+12x+8+9x^2-1=x^3+3x^2+3x+1\)
<=>\(9x+6=0\)
<=>x=\(\dfrac{-2}{3}\)
k. (x + 1)(2x – 3) = (2x – 1)(x + 5)
<=>\(2x^2-x-3=2x^2+9x-5\)
<=>10x=2
<=>x=\(\dfrac{1}{5}\)
\(\frac{2}{x^2-4}-\frac{x-1}{x\left(x-2\right)}+\frac{x-4}{x\left(x+2\right)}=0\)Đk \(x\ne\pm2;x\ne0\)
\(\Rightarrow\frac{2}{\left(x-2\right)\left(x+2\right)}-\frac{x-1}{x\left(x-2\right)}+\frac{x-4}{x\left(x+2\right)}=0\)
\(\Rightarrow\frac{2x-\left(x-1\right)\left(x+2\right)+\left(x-4\right)\left(x-2\right)}{x\left(x-2\right)\left(x+2\right)}=0\)
\(\Rightarrow2x-\left(x-1\right)\left(x+2\right)+\left(x-4\right)\left(x-2\right)=0\)
\(\Rightarrow2x-x^2-x+2+x^2-6x+8=0\)
\(\Rightarrow-5x+10=0\)
\(\Rightarrow-5x=-10\)
\(\Rightarrow x=2\)Loại
Ko có gt x thỏa mãn
\(\frac{1}{3-x}-\frac{1}{x+1}=\frac{x}{x-3}-\frac{\left(x-1\right)^2}{x^2-2x-3}\)
\(\Rightarrow\frac{1}{3-x}-\frac{1}{x+1}=\frac{x}{x-3}-\frac{\left(x-1\right)^2}{x^2-3x+x-3}\)
\(\Rightarrow\frac{1}{3-x}-\frac{1}{x+1}=\frac{x}{x-3}-\frac{\left(x-1\right)^2}{\left(x-3\right)\left(x+1\right)}\)Đk \(x\ne3;x\ne-1\)
\(\Leftrightarrow\frac{1}{3-x}-\frac{1}{x+1}-\frac{x}{x-3}-\frac{\left(x-1\right)^2}{\left(x-3\right)\left(x+1\right)}=0\)
\(\Rightarrow-\frac{1}{x-3}-\frac{1}{x+1}-\frac{x}{x-3}+\frac{\left(x-1\right)^2}{\left(x-3\right)\left(x+1\right)}=0\)
\(\Rightarrow\frac{-1\left(x+1\right)-1\left(x-3\right)-x\left(x+1\right)+\left(x-1\right)^2}{\left(x-3\right)\left(x+1\right)}=0\)
\(\Rightarrow-\left(x+1\right)-\left(x-3\right)-x\left(x+1\right)+\left(x-1\right)^2=0\)
\(\Rightarrow x-1-x+3-x^2-x+x^2-2x+1=0\)
\(\Rightarrow-3x+3=0\)
\(\Rightarrow-3x=-3\)
\(\Rightarrow x=1\)
a: \(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{3}{2}\end{matrix}\right.\)
b: \(\Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\\x=4\end{matrix}\right.\)
c: \(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\5x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{4}{5}\end{matrix}\right.\)
d: \(\Leftrightarrow\left(x+3\right)\left(x-4\right)=0\)
=>x+3=0 hoặc x-4=0
=>x=-3 hoặc x=4
e: \(\Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\\x=4\end{matrix}\right.\)
f: \(\Leftrightarrow\left(2x+3\right)\left(x-4\right)\left(x+4\right)=0\)
hay \(x\in\left\{-\dfrac{3}{2};4;-4\right\}\)
a, \(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{3}{2}\end{matrix}\right.\)
b, \(\Leftrightarrow\left[{}\begin{matrix}x^2-9=0\\4-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\pm3\\x=4\end{matrix}\right.\)
c, \(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\4-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{4}{5}\end{matrix}\right.\)
d, \(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=4\end{matrix}\right.\)
e, tương tự d
f, \(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\x^2-16=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\pm4\end{matrix}\right.\)
a: =>x^2+4x-4x+1=0
=>x^2+1=0
=>Loại
b: =>2x-6+4=2x+2
=>-2=2(loại)
c: =>2(x+3)-2x-1=1
=>6-1=1
=>5=1(loại)
d =>x+3=0
=>x=-3(loại)
e: =>x^2-3x^2+3x-3x-2=0
=>-2x^2-2=0
=>x^2+1=0
=>Loại
\(A=\left(3x-4\right)^2-4\left(x+1\right)^2=0\)
\(\Rightarrow\left(3x-4\right)^2-2^2\left(x+1\right)^2=0\)
\(\Rightarrow\left(3x-4\right)^2-\left(2x-2\right)^2=0\)
\(\Rightarrow\left(3x-4-2x+2\right)\left(3x-4+2x-2\right)=0\)
\(\Rightarrow\left(x-2\right)\left(5x-6\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\5x-6=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=\frac{6}{5}\end{cases}}\)