a) \(\sqrt[]{x^2-4x+4}=x+3\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-2=x+3\\x-2=-\left(x+3\right)\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}0x=5\left(loại\right)\\x-2=-x-3\end{matrix}\right.\)

\(\Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\)

b) \(2x^2-\sqrt[]{9x^2-6x+1}=5\)

\(\Leftrightarrow2x^2-\sqrt[]{\left(3x-1\right)^2}=5\)

\(\Leftrightarrow2x^2-\left|3x-1\right|=5\)

\(\Leftrightarrow\left|3x-1\right|=2x^2-5\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=2x^2-5\\3x-1=-2x^2+5\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2x^2-3x-4=0\left(1\right)\\2x^2+3x-6=0\left(2\right)\end{matrix}\right.\)

Giải pt (1)

\(\Delta=9+32=41>0\)

Pt \(\left(1\right)\) \(\Leftrightarrow x=\dfrac{3\pm\sqrt[]{41}}{4}\)

Giải pt (2)

\(\Delta=9+48=57>0\)

Pt \(\left(2\right)\) \(\Leftrightarrow x=\dfrac{-3\pm\sqrt[]{57}}{4}\)

Vậy nghiệm pt là \(\left[{}\begin{matrix}x=\dfrac{3\pm\sqrt[]{41}}{4}\\x=\dfrac{-3\pm\sqrt[]{57}}{4}\end{matrix}\right.\)

c: =>(x+2)(x+3)(x-5)(x-6)=180

=>(x^2-3x-10)(x^2-3x-18)=180

=>(x^2-3x)^2-28(x^2-3x)=0

=>x(x-3)(x-7)(x+4)=0

=>\(x\in\left\{0;3;7;-4\right\}\)

c: =>(x-3)(x+2)(2x+1)(3x-1)=0

=>\(x\in\left\{3;-2;-\dfrac{1}{2};\dfrac{1}{3}\right\}\)

Các bn giúp mình với mình đang cần gấp

nhiều quá bạn ơi , mk nghĩ bạn nên tách ra rồi hãy đăng lên

Bài 1:

16:

=>15-30x-8+8x=4x-9

=>-22x+7=4x-9

=>-26x=-16

=>x=8/13

15: \(\Leftrightarrow12x-6+12-4x=10-6x\)

=>8x+6=10-6x

=>14x=4

=>x=2/7

14: \(\Leftrightarrow12x-18+9-15x=8x-9\)

=>-3x-9=8x-9

=>x=0

13: \(\Leftrightarrow10x+15x-10=25-10x\)

=>25x-10=25-10x

=>35x=35

=>x=1

12: \(\Leftrightarrow15x-18-4x+10=11x-10\)

=>11x-8=11x-10(loại)

x²+10x+25-4x(x+5)=0

⇔(x+5)²-4x(x+5)=0

⇔(x+5)(x+5-4x)=0

⇔(x+5)(5-3x)=0

⇔\(\left\{{}\begin{matrix}x+5=0\\5-3x=0\end{matrix}\right.\Leftrightarrow\left\{{} }\left\{{}\begin{matrix}x=-5\\x=\dfrac{5}{3}\end{matrix}\right.\)

Ta có:

\(\sqrt{3x^2+6x+12}+\sqrt{5x^4-10x^2+9}=\sqrt{3\left(x+1\right)^2+9}+\sqrt{5\left(x^2-1\right)^2+4}\ge\sqrt{9}+\sqrt{4}=5\)

\(3-4x-2x^2=5-2\left(x+1\right)^2\le5\)

Đẳng thức xảy ra khi và chỉ khi:

\(\left\{{}\begin{matrix}3\left(x+1\right)^2=0\\5\left(x^2-1\right)^2=0\\2\left(x+1\right)^2=0\end{matrix}\right.\) \(\Rightarrow x=-1\)

Vậy pt có nghiệm duy nhất \(x=-1\)

\(a.\frac{4x-3}{x-5}=\frac{29}{3}\\ \Leftrightarrow\frac{3\left(4x-3\right)}{3\left(x-5\right)}=\frac{29\left(x-5\right)}{3\left(x-5\right)}\\ \Leftrightarrow3\left(4x-3\right)=29\left(x-5\right)\\ \Leftrightarrow3\left(4x-3\right)-29\left(x-5\right)=0\\ \Leftrightarrow12x-9-29x+145=0\\ \Leftrightarrow-17x+136=0\\ \Leftrightarrow-17x=-136\\ \Leftrightarrow x=\frac{-136}{-17}=8\)

\(b.\frac{2x-1}{5-3x}=2\\ \Leftrightarrow\frac{2x-1}{5-3x}=\frac{4}{2}\\ \Leftrightarrow\frac{2\left(2x-1\right)}{2\left(5-3x\right)}=\frac{4\left(5-3x\right)}{2\left(5-3x\right)}\\ \Leftrightarrow2\left(2x-1\right)=4\left(5-3x\right)\\ \Leftrightarrow2\left(2x-1\right)-4\left(5-3x\right)=0\\ \Leftrightarrow4x-2-20+12x=0\\ \Leftrightarrow16x-22=0\\ \Leftrightarrow16x=22\\ \Leftrightarrow x=\frac{22}{16}=\frac{11}{8}\)

\(c.\frac{4x-5}{x-1}=\frac{2+x}{x-1}\\ \Leftrightarrow4x-5=2+x\\ \Leftrightarrow4x-5-2-x=0\\ \Leftrightarrow3x-7=0\\ \Leftrightarrow3x=7\\ \Leftrightarrow x=\frac{7}{3}\)

\(d.\frac{7}{x+2}=\frac{3}{x-5}\\ \Leftrightarrow\frac{7\left(x-5\right)}{\left(x+2\right)\left(x-5\right)}=\frac{3\left(x+2\right)}{\left(x+2\right)\left(x-5\right)}\\ \Leftrightarrow7\left(x-5\right)=3\left(x+2\right)\\ \Leftrightarrow7\left(x-5\right)-3\left(x+2\right)=0\\ \Leftrightarrow7x-35-3x-6=0\\ \Leftrightarrow4x-41=0\\ \Leftrightarrow4x=41\\ \Leftrightarrow x=\frac{41}{4}\)

\(e.\frac{2x+5}{2x}-\frac{x}{x+5}=0\\ \Leftrightarrow\frac{\left(2x+5\right)\left(x+5\right)}{2x\left(x+5\right)}-\frac{x.2x}{2x\left(x+5\right)}=0\\ \Leftrightarrow\left(2x+5\right)\left(x+5\right)-2x^2=0\\ \Leftrightarrow2x^2+10x+5x+25-2x^2=0\\ \Leftrightarrow15x+25=0\\ \Leftrightarrow15x=-25\\ \Leftrightarrow x=\frac{-25}{15}=\frac{-5}{3}\)

\(f.\frac{12x+1}{11x-4}+\frac{10x-4}{9}=\frac{20x+17}{18}\\\Leftrightarrow\frac{18\left(12x+1\right)}{18\left(11x-4\right)}+\frac{\left(10x-4\right).2\left(11x-4\right)}{9.2\left(11x-4\right)}=\frac{\left(20x+17\right)\left(11x-4\right)}{18\left(11x-4\right)}\\ \Leftrightarrow18\left(12x+1\right)+\left(10x-4\right).2\left(11x-4\right)=\left(20x+17\right)\left(11x-4\right)\\ \Leftrightarrow220x^2+48x+50=220x^2+107x-68\\ \Leftrightarrow48x+50=107x-68\\ \Leftrightarrow48x-107x=-68-50\\ \Leftrightarrow59x=-118\\ \Leftrightarrow x=-2\)

1) \(\sqrt{5-2x}=6\left(đk:x\le\dfrac{5}{2}\right)\)

\(\Leftrightarrow5-2x=36\)

\(\Leftrightarrow2x=-31\Leftrightarrow x=-\dfrac{31}{2}\left(tm\right)\)

2) \(\sqrt{2-x}=\sqrt{x+1}\left(đk:2\ge x\ge-1\right)\)

\(\Leftrightarrow2-x=x+1\)

\(\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\left(tm\right)\)

3) \(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)

\(\Leftrightarrow\left|2x+1\right|=6\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

4) \(\sqrt{x^2-10x+25}=x-2\left(đk:x\ge2\right)\)

\(\Leftrightarrow\sqrt{\left(x-5\right)^2}=x-2\)

\(\Leftrightarrow\left|x-5\right|=x-2\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=x-2\left(x\ge5\right)\\x-5=2-x\left(2\le x< 5\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}5=2\left(VLý\right)\\x=\dfrac{7}{2}\left(tm\right)\end{matrix}\right.\)