Giải phương trình \(\sqrt{4x^2+5x+1}-2\sqrt{x^2-1+1}=3-9x\)
Giải phương trình: \(\sqrt{4x^2+5x+1}-9x=2\sqrt{x^2-x+1}-3\)
Giải phương trình: \(\sqrt{4x^2+5x+1}-2\sqrt{x^2-x+1}=3-9x\)
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
\(\sqrt{4x^2-4x+1}=3-x\)
\(\sqrt{9x+9}+\sqrt{x+1}-\sqrt{4x+4}=2\left(x+1\right)\)
\(\sqrt{4x^2-4x+1}=3-x\left(x\in R\right)\\ \Leftrightarrow\sqrt{\left(2x-1\right)^2}=3-x\\ \Leftrightarrow2x-1=3-x\\ \Leftrightarrow3x=4\Leftrightarrow x=\dfrac{4}{3}\\ \sqrt{9x+9}+\sqrt{x+1}-\sqrt{4x+4}=2\left(x+1\right)\left(x\ge-1\right)\\ \Leftrightarrow\sqrt{x+1}\left(\sqrt{9}+1+\sqrt{4}\right)=2\left(x+1\right)\\ \Leftrightarrow6\sqrt{x+1}=2\left(x+1\right)\\ \Leftrightarrow3\sqrt{x+1}=x+1\\ \Leftrightarrow\sqrt{x+1}\left(3-\sqrt{x+1}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+1=0\\\sqrt{x+1}=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x+1=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\left(tm\right)\\x=8\left(tm\right)\end{matrix}\right.\)
a, ĐK: \(x\in R\)
\(\sqrt{4x^2-4x+1}=3-x\)
\(\Leftrightarrow\sqrt{\left(2x-1\right)^2}=3-x\)
\(\Leftrightarrow\left|2x-1\right|=3-x\)
TH1: \(\left\{{}\begin{matrix}2x-1\ge0\\2x-1=3-x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{2}\\x=\dfrac{4}{3}\end{matrix}\right.\Leftrightarrow x=\dfrac{4}{3}\)
TH2: \(\left\{{}\begin{matrix}2x-1< 0\\1-2x=3-x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< \dfrac{1}{2}\\x=-2\end{matrix}\right.\Leftrightarrow x=-2\)
b, ĐK: \(x\ge-1\)
\(\sqrt{9x+9}+\sqrt{x+1}-\sqrt{4x+4}=2\left(x+1\text{}\right)\)
\(\Leftrightarrow3\sqrt{x+1}+\sqrt{x+1}-2\sqrt{x+1}=2\left(x+1\right)\)
\(\Leftrightarrow\sqrt{x+1}=x+1\)
\(\Leftrightarrow\sqrt{x+1}\left(\sqrt{x+1}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=0\\\sqrt{x+1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\left(tm\right)\\x=0\left(tm\right)\end{matrix}\right.\)
GIẢI PHƯƠNG TRÌNH
a) \(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9x-18}+6\sqrt{\dfrac{x-2}{81}}=-4\)
b) \(\sqrt{9x^2+12x+4}=4x\)
c) \(\sqrt{9x-18}-\sqrt{4x-8}+3\sqrt{x-2}=40\)
d) \(\sqrt{5x-6}-3=0\)
a: \(\Leftrightarrow\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\cdot3\sqrt{x-2}+6\cdot\dfrac{\sqrt{x-2}}{9}=-4\)
\(\Leftrightarrow\sqrt{x-2}=4\)
=>x-2=16
hay x=18
b: \(\Leftrightarrow\left|3x+2\right|=4x\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+2=4x\left(x>=-\dfrac{2}{3}\right)\\3x+2=-4x\left(x< -\dfrac{2}{3}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(nhận\right)\\x=-\dfrac{2}{7}\left(nhận\right)\end{matrix}\right.\)
c: \(\Leftrightarrow3\sqrt{x-2}-2\sqrt{x-2}+3\sqrt{x-2}=40\)
\(\Leftrightarrow4\sqrt{x-2}=40\)
=>x-2=100
hay x=102
d: =>5x-6=9
hay x=3
\(a,\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9x-18}+6\sqrt{\dfrac{x-2}{81}}=-4\left(dk:x\ge2\right)\)
\(\Leftrightarrow\dfrac{1}{3}\sqrt{x-2}-2\sqrt{x-2}+\dfrac{2}{3}\sqrt{x-2}=-4\)
\(\Leftrightarrow\sqrt{x-2}=4\)
\(\Leftrightarrow x-2=16\)
\(\Leftrightarrow x=18\left(tmdk\right)\)
b,\(\sqrt{9x^2-12x+4=3x\left(dk:x\ge0\right)}\)
\(\Leftrightarrow\sqrt{\left(3x-2\right)^2}=3x\)
\(\Leftrightarrow\left|3x-2\right|=3x\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x-2=3x\\3x-2=-3x\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x\in\varnothing\\x=\dfrac{1}{3}\left(tmdk\right)\end{matrix}\right.\)
Các câu còn lại làm tương tự nhé
\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9x-18}+6\sqrt{\dfrac{x-2}{81}}=-4\) (đk: x≥2)
\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9\left(x-2\right)}+6\sqrt{\dfrac{1}{81}\left(x-2\right)}=-4\)
\(\dfrac{1}{3}\sqrt{x-2}-2\sqrt{x-2}+\dfrac{2}{3}\sqrt{x-2}=-4\)
\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{4}{3}\sqrt{x-2}=-4\)
\(-\sqrt{x-2}=-4\)
\(\sqrt{x-2}=4\)
\(\left|x-2\right|=16\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=16\\x-2=-16\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=18\left(TM\right)\\x=-14\left(L\right)\end{matrix}\right.\)
giải phương trình với điều kiện x thuộc R: \(\sqrt{4x^2+5x-1}-2\sqrt{x^2-x-1}=9x+3\)
nguồn: Bad Luck t2 :D
Đặt \(\sqrt{4x^2+5x-1}=a;2\sqrt{x^2-x-1}=b\left(a\ge0,b\ge0\right)\Rightarrow a^2-b^2=9x+3\)
Ta thụ được hệ phương trình: \(\hept{\begin{cases}a^2-b^2=9x+3\\a-b=9x+3\end{cases}\Rightarrow a^2-b^2=a-b\Leftrightarrow\left(a-b\right)\left(a+b-1\right)=0\Leftrightarrow\orbr{\begin{cases}a=b\\a+b=1\end{cases}}}\)
Xét 2 trường hợp xảy ra:
TH1: \(a=b\Leftrightarrow9x+3=0\Leftrightarrow x=\frac{-1}{3}\left(lo\text{ại}\right)\)
TH2: Kết hợp \(\hept{\begin{cases}a+b=1\\a-b=9x+3\end{cases}\Rightarrow2a=9x+4\Leftrightarrow\hept{\begin{cases}x\ge\frac{-4}{9}\\4\left(4x^2+5x-1\right)=81x^2+72x+16\end{cases}}}\)\(\Leftrightarrow\hept{\begin{cases}x\ge\frac{-4}{9}\\65x^2+52x+20=0\end{cases}}\)(*)
Hệ điều kiện (*) vô nghiệ do phương trình \(65x^2+52x+20=0\)vô nghiệm
Vậy hệ phương trình đã cho vô nghiệm.
đk: \(\orbr{\begin{cases}x\ge\frac{1+\sqrt{5}}{2}\\x\le\frac{-5-\sqrt{41}}{8}\end{cases}}\)
Đặt \(\hept{\begin{cases}\sqrt{4x^2+5x-1}=a\\\sqrt{x^2-x-1}=b\end{cases}}\Leftrightarrow\hept{\begin{cases}4x^2+5x-1=a^2\\4\left(x^2-x-1\right)=4b^2\end{cases}}\)
\(\Rightarrow a^2-4b^2=9x+3\)
Mà \(a-2b=9x+3\)
=> \(a^2-4b^2=a-2b\)
<=> \(\left(a-2b\right)\left(a+2b\right)-\left(a-2b\right)=0\)
<=> \(\left(a-2b\right)\left(a+2b-1\right)=0\)
<=> \(\orbr{\begin{cases}a-2b=0\\a+2b-1=0\end{cases}}\)
Nếu: \(a-2b=0\)
\(\Leftrightarrow9x+3=0\)
\(\Leftrightarrow9x=-3\)
\(\Rightarrow x=-\frac{1}{3}\left(tm\right)\)
Nếu: \(a+2b-1=0\)
\(\Rightarrow a+2b=1\) , mà \(a-2b=9x+3\)
=> \(2a=9x+4\)
<=> \(2\sqrt{4x^2+5x-1}=9x+4\)
<=> \(4\left(4x^2+5x-1\right)=81x^2+72x+16\)
<=> \(65x^2+52x+20=0\)
<=> \(65\left(x^2+\frac{4}{5}x+\frac{4}{25}\right)+\frac{48}{5}=0\)
\(\Leftrightarrow65\left(x+\frac{2}{5}\right)^2=-\frac{48}{5}\) (vô lý)
Vậy \(x=-\frac{1}{3}\)
Theo quan điểm cá nhân là vậy._.
Sorry nhầm đoạn \(x=-\frac{1}{3}\) không thỏa mãn
=> PT vô nghiệm
giải các phương trình sau :
a.\(\sqrt{2x-2}-\sqrt{6x-9}=16x^2-48x+35\)
b.\(\sqrt{4x^2+5x+1}-2\sqrt{x^2-x+1}=9x-3\)
Giải phương trình sau
a)\(\sqrt{4x^2+5x+1}-2\sqrt{x^2-x+1}=9x-3\)
b) \(\sqrt{2x-2}-\sqrt{6x-9}=16x^2-48x+35\)
a/ Điều kiện b tự làm nhé
Đặt \(\hept{\begin{cases}\sqrt{4x^2+5x+1}=a\left(a\ge0\right)\\2\sqrt{x^2-x+1}=b\left(b\ge0\right)\end{cases}}\)
Ta có: \(a^2-b^2=9x-3\)từ đó pt ban đầu thành
\(a-b=a^2-b^2\)
\(\Leftrightarrow\left(a-b\right)\left(1-a-b\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=b\\1=a+b\end{cases}}\)
Tới đây thì đơn giản rồi b làm tiếp nhé
Giai phương trình \(\sqrt{4x^2+5x+1}-2\sqrt{x^2-x+1}=9x-3\)
ĐKXĐ : \(4x^2+5x+1\ge0\Leftrightarrow\left(4x+1\right)\left(x+1\right)\ge0\Rightarrow\orbr{\begin{cases}x\le-1\\x\ge-\frac{1}{4}\end{cases}}\)
\(\sqrt{4x^2+5x+1}-2\sqrt{x^2-x+1}=9x-3\)
\(\Leftrightarrow\sqrt{4x^2+5x+1}-\frac{2\sqrt{7}}{3}-2\sqrt{x^2-x+1}+\frac{2\sqrt{7}}{3}-9x+3=0\)
\(\Leftrightarrow\frac{4x^2+5x+1-\frac{28}{9}}{\sqrt{4x^2+5x+1}+\frac{2\sqrt{7}}{3}}-2\left(\frac{x^2-x+1-\frac{7}{9}}{\sqrt{x^2-x+1}+\frac{\sqrt{7}}{3}}\right)+3\left(3x-1\right)=0\)
\(\Leftrightarrow\frac{4x^2+5x-\frac{19}{9}}{\sqrt{4x^2+5x+1}+\frac{2\sqrt{7}}{3}}-2.\frac{x^2-x+\frac{2}{9}}{\sqrt{x^2-x+1}+\frac{\sqrt{7}}{3}}+3\left(3x-1\right)=0\)
\(\Leftrightarrow\frac{\left(x-\frac{1}{3}\right)\left(4x+\frac{19}{3}\right)}{\sqrt{4x^2+5x+1}+\frac{2\sqrt{7}}{3}}-\frac{2\left(x-\frac{2}{3}\right)\left(x-\frac{1}{3}\right)}{\sqrt{x^2-x+1}+\frac{\sqrt{7}}{3}}+9\left(x-\frac{1}{3}\right)=0\)
\(\Leftrightarrow\left(x-\frac{1}{3}\right)\left(\frac{4x+\frac{19}{3}}{\frac{2\sqrt{7}}{3}}-\frac{2x-\frac{4}{3}}{\sqrt{x^2-x+1}+\frac{\sqrt{7}}{2}}+9\right)=0\)
\(\Rightarrow x=\frac{1}{3}\)(TMĐKXĐ)
giải phương trình
a)\(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\)
b)\(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=16\)
c)\(\sqrt{4x+20}+\sqrt{x+5}-\dfrac{1}{3}\sqrt{9x+45}=4\)
d)\(\dfrac{1}{3}\sqrt{2x}-\sqrt{8x}+\sqrt{18x}-10=2\)
a) \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\) (ĐK: \(x\ge1\))
\(\Leftrightarrow\sqrt{x-1}+\sqrt{4\left(x-1\right)}-\sqrt{25\left(x-1\right)}+2=0\)
\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)
\(\Leftrightarrow-2\sqrt{x-1}=-2\)
\(\Leftrightarrow\sqrt{x-1}=\dfrac{2}{2}\)
\(\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\)
\(\Leftrightarrow x=2\left(tm\right)\)
b) \(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=16\) (ĐK: \(x\ge-1\))
\(\Leftrightarrow\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}+\sqrt{4\left(x+1\right)}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}=16\)
\(\Leftrightarrow\sqrt{x+1}=4\)
\(\Leftrightarrow x+1=16\)
\(\Leftrightarrow x=15\left(tm\right)\)