Giải phương trình \(x^2+\dfrac{9x^2}{\left(x+3\right)^2}=40\)
Giải hệ phương trình : \(\left\{{}\begin{matrix}x^3-2y^2-12y=26\\x^2y^2+9x^2-3y^2-6y=27\end{matrix}\right.\)
Giải phương trình : \(\left(\dfrac{x}{x+2}\right)^2=3x^2-6x-3\)
\(\sqrt{x+5}=x^2-5\)
\(\sqrt{x-1}-\sqrt[3]{2-x}=5\)
2b. ĐKXĐ : \(x\ge-5\) (*)
Ta có \(\sqrt{x+5}=x^2-5\)
\(\Leftrightarrow4x^2-20-4\sqrt{x+5}=0\)
\(\Leftrightarrow4x^2+4x+1-4.\left(x+5\right)-4\sqrt{x+5}-1=0\)
\(\Leftrightarrow\left(2x+1\right)^2-\left(2\sqrt{x+5}+1\right)^2=0\)
\(\Leftrightarrow\left(x+1+\sqrt{x+5}\right)\left(x-\sqrt{x+5}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=-\sqrt{x+5}\left(1\right)\\x=\sqrt{x+5}\left(2\right)\end{matrix}\right.\)
Giải (1) có (1) \(\Leftrightarrow\left(x+1\right)^2=x+5\) ; ĐK: \(\left(x\le-1\right)\)
\(\Leftrightarrow x^2+x-4=0\Leftrightarrow x=\dfrac{-1\pm\sqrt{17}}{2}\)
Kết hợp (*) và ĐK được \(x=\dfrac{-1-\sqrt{17}}{2}\) là nghiệm phương trình gốc
Giải (2) có (2) <=> \(x^2-x-5=0\) ; ĐK : \(x\ge0\)
\(\Leftrightarrow x=\dfrac{1\pm\sqrt{21}}{2}\)
Kết hợp (*) và ĐK được \(x=\dfrac{1+\sqrt{21}}{2}\) là nghiệm phương trình gốc
Tập nghiệm \(S=\left\{\dfrac{-1-\sqrt{17}}{2};\dfrac{1+\sqrt{21}}{2}\right\}\)
2c. ĐKXĐ \(x\ge1\) (*)
Đặt \(\sqrt{x-1}=a;\sqrt[3]{2-x}=b\left(a\ge0\right)\) (1)
Ta có \(\sqrt{x-1}-\sqrt[3]{2-x}=5\Leftrightarrow a-b=5\)
Từ (1) có \(a^2+b^3=1\) (2)
Thế a = b + 5 vào (2) ta được
\(b^3+\left(b+5\right)^2=1\Leftrightarrow b^3+b^2+10b+24=0\)
\(\Leftrightarrow b^3+8+b^2+10b+16=0\)
\(\Leftrightarrow\left(b+2\right).\left(b^2-b+12\right)=0\)
\(\Leftrightarrow b=-2\) (Vì \(b^2-b+12=\left(b-\dfrac{1}{2}\right)^2+\dfrac{47}{4}>0\forall b\)
Với b = -2 \(\Leftrightarrow\sqrt[3]{2-x}=-2\Leftrightarrow x=10\) (tm)
Tập nghiệm \(S=\left\{10\right\}\)
Giải phương trình sau:
a) \(\sqrt{4x+20}-3\sqrt{5+x}+\dfrac{4}{3}\sqrt{9x+45}=6\)
b) \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)
c) \(2x-x^2+\sqrt{6x^2-12x+7}=0\)
d) \(\left(x+1\right)\left(x+4\right)-3\sqrt{x^2+5x+2}=6\)
a: Ta có: \(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45}=6\)
\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)
\(\Leftrightarrow3\sqrt{x+5}=6\)
\(\Leftrightarrow x+5=4\)
hay x=-1
b: Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)
\(\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)
\(\Leftrightarrow\sqrt{x-1}=17\)
\(\Leftrightarrow x-1=289\)
hay x=290
giải phương trình 1)\(\dfrac{1-6x}{x-2}+\dfrac{9x+4}{x+2}=\dfrac{x\left(3x-2\right)+1}{x^2-4}\)2) \(\dfrac{3x+2}{3x-2}-\dfrac{6}{2+3x}=\dfrac{9x^2}{9x^2-4}\)3) \(\dfrac{x+5}{3x-6}-\dfrac{1}{2}=\dfrac{2x-3}{2x-4}\)4) \(\dfrac{x-1}{x}+\dfrac{1}{x+1}=\dfrac{2x-1}{2x^2+2}\)5) \(\dfrac{2}{x+1}+\dfrac{3x+1}{x+1}=\dfrac{1}{\left(x+1\right)\left(x-2\right)}\)
giúp mình với ạ câu nào cũng được
tìm n của phương trình \(x^2-\dfrac{2n-2x}{4}-2x+5n=x^3-9x^2+10\)
có nghiệm bằng \(\dfrac{1}{3}\)của phương trình \(\left(x+1\right)\left(x+3\right)=x\left(x-3\right)+24\)
Giải phương trình
1, \(\dfrac{1-6x}{x-2}+\dfrac{9x+4}{x+2}=\dfrac{x\left(3x-2\right)+1}{x^2-4}\)
2, \(\dfrac{3x+2}{3x-2}-\dfrac{6}{2-3x}=\dfrac{9x^2}{9x^2-4}\)3, \(\dfrac{x-1}{x}+\dfrac{1}{x+1}=\dfrac{2x-1}{2x^2+2}\)4, \(\dfrac{2}{x+1}+\dfrac{3x+1}{x+1}=\dfrac{1}{\left(x+1\right)\left(x-2\right)}\)5, \(\dfrac{x+5}{3x-6}-\dfrac{1}{2}=\dfrac{2x-3}{2x-4}\)
1) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
Ta có: \(\dfrac{1-6x}{x-2}+\dfrac{9x+4}{x+2}=\dfrac{x\left(3x-2\right)+1}{x^2-4}\)
\(\Leftrightarrow\dfrac{\left(1-6x\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(9x+4\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{3x^2-2x+1}{\left(x-2\right)\left(x+2\right)}\)
Suy ra: \(\left(1-6x\right)\left(x+2\right)+\left(9x+4\right)\left(x-2\right)=3x^2-2x+1\)
\(\Leftrightarrow x+2-6x^2-12x+9x^2-18x+4x-8-3x^2+2x-1=0\)
\(\Leftrightarrow-23x-7=0\)
\(\Leftrightarrow-23x=7\)
\(\Leftrightarrow x=-\dfrac{7}{23}\)(nhận)
Vậy: \(S=\left\{-\dfrac{7}{23}\right\}\)
2) ĐKXĐ: \(x\notin\left\{\dfrac{2}{3};-\dfrac{2}{3}\right\}\)
Ta có: \(\dfrac{3x+2}{3x-2}-\dfrac{6}{2-3x}=\dfrac{9x^2}{9x^2-4}\)
\(\Leftrightarrow\dfrac{3x+2}{3x-2}+\dfrac{6}{3x-2}=\dfrac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)
\(\Leftrightarrow\dfrac{3x+8}{3x-2}=\dfrac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)
\(\Leftrightarrow\dfrac{\left(3x+8\right)\left(3x+2\right)}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)
Suy ra: \(9x^2+6x+24x+16=9x^2\)
\(\Leftrightarrow30x+16=0\)
\(\Leftrightarrow30x=-16\)
hay \(x=-\dfrac{8}{15}\)(nhận)
Vậy: \(S=\left\{-\dfrac{8}{15}\right\}\)
Giải Phương Trình
\(\sqrt{\left(2x+3\right)^2}=5\)
\(\sqrt{9\left(x-2\right)^2}=18\)
\(\sqrt{9x-18}-\sqrt{4x-8}+3\sqrt{x-2}=40\)
\(\sqrt{4.\left(x-3\right)^2}=8\)
\(\sqrt{5x-6}-3=0\)
Giải các phương trình sau:
a. \(\sqrt{25x+75}+3\sqrt{x-2}=2\sqrt{x-2}+\sqrt{9x-18}\)
b. \(\sqrt{\left(2x-1\right)^2}=4\)
c. \(\sqrt{\left(2x+1\right)^2}=3x-5\)
d. \(\sqrt{4x-12}-14\sqrt{\dfrac{x-2}{49}}=\sqrt{9x-18}+8\)
a) Ta có: \(\sqrt{25x+75}+3\sqrt{x-2}=2\sqrt{x-2}+\sqrt{9x-18}\)
\(\Leftrightarrow5\sqrt{x+3}+3\sqrt{x-2}=2\sqrt{x-2}+3\sqrt{x-2}\)
\(\Leftrightarrow\sqrt{25x+75}=\sqrt{4x-8}\)
\(\Leftrightarrow25x-4x=-8-75\)
\(\Leftrightarrow21x=-83\)
hay \(x=-\dfrac{83}{21}\)
b) Ta có: \(\sqrt{\left(2x-1\right)^2}=4\)
\(\Leftrightarrow\left|2x-1\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=4\\2x-1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
c) Ta có: \(\sqrt{\left(2x+1\right)^2}=3x-5\)
\(\Leftrightarrow\left|2x+1\right|=3x-5\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=3x-5\left(x\ge-\dfrac{1}{2}\right)\\2x+1=5-3x\left(x< \dfrac{1}{2}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3x=-5-1\\2x+3x=5-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\left(nhận\right)\\x=\dfrac{4}{5}\left(loại\right)\end{matrix}\right.\)
d) Ta có: \(\sqrt{4x-12}-14\sqrt{\dfrac{x-2}{49}}=\sqrt{9x-18}+8\)
\(\Leftrightarrow2\sqrt{x-3}-2\sqrt{x-2}=3\sqrt{x-2}+8\)
\(\Leftrightarrow2\sqrt{x-3}-5\sqrt{x-2}=8\)
\(\Leftrightarrow4\left(x-3\right)+25\left(x-2\right)-20\sqrt{x^2-5x+6}=8\)
\(\Leftrightarrow4x-12+25x-50-8=20\sqrt{\left(x-2\right)\left(x-3\right)}\)
\(\Leftrightarrow20\sqrt{\left(x-2\right)\left(x-3\right)}=29x-70\)
\(\Leftrightarrow x^2-5x+6=\dfrac{\left(29x-70\right)^2}{400}\)
\(\Leftrightarrow x^2-5x+6=\dfrac{841}{400}x^2-\dfrac{203}{20}x+\dfrac{49}{4}\)
\(\Leftrightarrow\dfrac{-441}{400}x^2+\dfrac{103}{20}x-\dfrac{25}{4}=0\)
\(\Delta=\left(\dfrac{103}{20}\right)^2-4\cdot\dfrac{-441}{400}\cdot\dfrac{-25}{4}=-\dfrac{26}{25}\)(Vô lý)
vậy: Phương trình vô nghiệm
tính đạo hàm
a) \(y=\dfrac{\left(x-2\right)^2}{\left(2x-3\right)\left(x-1\right)}\)
b) \(y=x+3+\dfrac{4}{x+3}\) giải phương trình y'=0
c) \(y=\dfrac{\left(5x-1\right)\left(x+1\right)}{x+2}\) tính y'(-1)
d) \(y=x-2+\dfrac{9}{x-2}\) giải phương trình y'=0
a:
ĐKXĐ: \(x\notin\left\{\dfrac{3}{2};1\right\}\)
\(y=\dfrac{\left(x-2\right)^2}{\left(2x-3\right)\left(x-1\right)}=\dfrac{x^2-4x+4}{2x^2-2x-3x+3}\)
=>\(y=\dfrac{x^2-4x+4}{2x^2-5x+3}\)
=>\(y'=\dfrac{\left(x^2-4x+4\right)'\left(2x^2-5x+3\right)-\left(x^2-4x+4\right)\left(2x^2-5x+3\right)'}{\left(2x^2-5x+3\right)^2}\)
=>\(y'=\dfrac{\left(2x-4\right)\left(2x^2-5x+3\right)-\left(2x-5\right)\left(x^2-4x+4\right)}{\left(2x^2-5x+3\right)^2}\)
=>\(y'=\dfrac{4x^3-10x^2+6x-8x^2+20x-12-2x^3+8x^2-8x+5x^2-20x+20}{\left(2x^2-5x+3\right)^2}\)
=>\(y'=\dfrac{2x^3-5x^2-2x+8}{\left(2x^2-5x+3\right)^2}\)
b:
ĐKXĐ: x<>-3
\(y=\left(x+3\right)+\dfrac{4}{x+3}\)
=>\(y'=\left(x+3+\dfrac{4}{x+3}\right)'=1+\left(\dfrac{4}{x+3}\right)'\)
\(=1+\dfrac{4'\left(x+3\right)-4\left(x+3\right)'}{\left(x+3\right)^2}\)
=>\(y'=1+\dfrac{-4}{\left(x+3\right)^2}=\dfrac{\left(x+3\right)^2-4}{\left(x+3\right)^2}\)
y'=0
=>\(\left(x+3\right)^2-4=0\)
=>\(\left(x+3+2\right)\left(x+3-2\right)=0\)
=>(x+5)(x+1)=0
=>x=-5 hoặc x=-1
c:
ĐKXĐ: x<>-2
\(y=\dfrac{\left(5x-1\right)\left(x+1\right)}{x+2}\)
=>\(y=\dfrac{5x^2+5x-x-1}{x+2}=\dfrac{5x^2+4x-1}{x+2}\)
=>\(y'=\dfrac{\left(5x^2+4x-1\right)'\left(x+2\right)-\left(5x^2+4x-1\right)\left(x+2\right)'}{\left(x+2\right)^2}\)
=>\(y'=\dfrac{\left(5x+4\right)\left(x+2\right)-\left(5x^2+4x-1\right)}{\left(x+2\right)^2}\)
=>\(y'=\dfrac{5x^2+10x+4x+8-5x^2-4x+1}{\left(x+2\right)^2}\)
=>\(y'=\dfrac{10x+9}{\left(x+2\right)^2}\)
\(y'\left(-1\right)=\dfrac{10\cdot\left(-1\right)+9}{\left(-1+2\right)^2}=\dfrac{-1}{1}=-1\)
d:
ĐKXĐ: x<>2
\(y=x-2+\dfrac{9}{x-2}\)
=>\(y'=\left(x-2+\dfrac{9}{x-2}\right)'=1+\left(\dfrac{9}{x-2}\right)'\)
\(=1+\dfrac{9'\left(x-2\right)-9\left(x-2\right)'}{\left(x-2\right)^2}\)
=>\(y'=1+\dfrac{-9}{\left(x-2\right)^2}=\dfrac{\left(x-2\right)^2-9}{\left(x-2\right)^2}\)
y'=0
=>\(\dfrac{\left(x-2\right)^2-9}{\left(x-2\right)^2}=0\)
=>\(\left(x-2\right)^2-9=0\)
=>(x-2-3)(x-2+3)=0
=>(x-5)(x+1)=0
=>x=5 hoặc x=-1
giải phương trình :
4x3+12x2+9x= -\(-\left(\dfrac{x+3}{x+1}\right)^3+6\left(\dfrac{x+3}{x+1}\right)+5\)