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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
a: ĐKXĐ: x>=-3/2
\(\sqrt{x^2+4}=\sqrt{2x+3}\)
=>\(x^2+4=2x+3\)
=>\(x^2-2x+1=0\)
=>\(\left(x-1\right)^2=0\)
=>x-1=0
=>x=1(nhận)
b: \(\sqrt{x^2-6x+9}=2x-1\)(ĐKXĐ: \(x\in R\))
=>\(\sqrt{\left(x-3\right)^2}=2x-1\)
=>\(\left\{{}\begin{matrix}\left(2x-1\right)^2=\left(x-3\right)^2\\x>=\dfrac{1}{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(2x-1-x+3\right)\left(2x-1+x-3\right)=0\\x>=\dfrac{1}{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(x+2\right)\left(3x-4\right)=0\\x>=\dfrac{1}{2}\end{matrix}\right.\)
=>x=4/3(nhận) hoặc x=-2(loại)
c:
Sửa đề: \(\sqrt{4x+12}=\sqrt{9x+27}-5\)
ĐKXĐ: \(x>=-3\)
\(\sqrt{4x+12}=\sqrt{9x+27}-5\)
=>\(2\sqrt{x+3}=3\sqrt{x+3}-5\)
=>\(-\sqrt{x+3}=-5\)
=>x+3=25
=>x=22(nhận)
d: ĐKXĐ: \(\left[{}\begin{matrix}x< =\dfrac{3-\sqrt{5}}{4}\\x>=\dfrac{3+\sqrt{5}}{4}\end{matrix}\right.\)
\(\sqrt{4x^2-6x+1}=\left|2x-5\right|\)
=>\(\sqrt{\left(4x^2-6x+1\right)}=\sqrt{4x^2-20x+25}\)
=>\(4x^2-6x+1=4x^2-20x+25\)
=>\(-6x+20x=25-1\)
=>\(14x=24\)
=>x=12/7(nhận)
a: ĐKXĐ: \(\left[{}\begin{matrix}x>=2\\x< =-3\end{matrix}\right.\)
\(\sqrt{\left(x-2\right)\left(x+3\right)}=5\)
=>\(\sqrt{x^2+x-6}=5\)
=>\(x^2+x-6=25\)
=>\(x^2+x-31=0\)
=>\(\left[{}\begin{matrix}x=\dfrac{-1+5\sqrt{5}}{2}\left(nhận\right)\\x=\dfrac{-1-5\sqrt{5}}{2}\left(nhận\right)\end{matrix}\right.\)
b: ĐKXĐ: \(x\in R\)
\(\sqrt{\left(2x+3\right)^2}=x-5\)
=>\(\left|2x+3\right|=x-5\)
=>\(\left\{{}\begin{matrix}x>=5\\\left(2x+3\right)^2=\left(x-5\right)^2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=5\\\left(2x+3-x+5\right)\left(2x+3+x-5\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=5\\\left(x+8\right)\left(3x-2\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=5\\\left[{}\begin{matrix}x=-8\left(loại\right)\\x=\dfrac{2}{3}\left(loại\right)\end{matrix}\right.\end{matrix}\right.\)
=>\(x\in\varnothing\)
c: ĐKXĐ: \(x\in R\)
\(\sqrt{x^2-6x+9}=x+7\)
=>\(\sqrt{\left(x-3\right)^2}=x+7\)
=>\(\left|x-3\right|=x+7\)
=>\(\left\{{}\begin{matrix}x+7>=0\\\left(x-3\right)^2=\left(x+7\right)^2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=-7\\\left(x-3-x-7\right)\left(x-3+x+7\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=-7\\-10\left(2x+4\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=-7\\x+2=0\end{matrix}\right.\)
=>x=-2
d: ĐKXĐ: x>=3/2
\(\sqrt{2x-3}=x-1\)
=>\(\left\{{}\begin{matrix}2x-3=\left(x-1\right)^2\\x>=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x^2-2x+1=2x-3\\x>=\dfrac{3}{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x^2-4x+4=0\\x>=\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x-2\right)^2=0\\x>=\dfrac{3}{2}\end{matrix}\right.\)
=>x=2
a) \(\sqrt{x}-x-0\) (ĐK: \(x\ge0\))
\(\Leftrightarrow\sqrt{x}\left(1-\sqrt{x}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=0\\1-\sqrt{x}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\sqrt{x}=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=1\left(tm\right)\end{matrix}\right.\)
b) \(x-\sqrt{2x-9}=6\)
\(\Leftrightarrow\sqrt{2x-9}=x-6\) (ĐK: \(x\ge\dfrac{9}{2}\))
\(\Leftrightarrow2x-9=\left(x-6\right)^2\)
\(\Leftrightarrow2x-9=x^2-12x+36\)
\(\Leftrightarrow x^2-14x+45=0\)
\(\Leftrightarrow x^2-5x-9x+45=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=9\left(tm\right)\end{matrix}\right.\)
c) \(3x-\sqrt{6x-\left(3-2\right)}=0\) (ĐK: \(x\ge\dfrac{1}{6}\))
\(\Leftrightarrow3x-\sqrt{6x-1}=0\)
\(\Leftrightarrow\sqrt{6x-1}=3x\)
\(\Leftrightarrow6x-1=9x^2\)
\(\Leftrightarrow9x^2-6x+1=0\)
\(\Leftrightarrow\left(3x-1\right)^2=0\)
\(\Leftrightarrow x=\dfrac{1}{3}\left(tm\right)\)
\(\Leftrightarrow-\left(x^2-2x\right)+\sqrt{6\left(x^2-2x\right)+7}=0\) ĐK \(\sqrt{6x^2-12x+7}\ge0\)
Đặt \(t=x^2-2x\left(t\ge0\right)\Leftrightarrow pt:-t+\sqrt{6t+7}=0\Leftrightarrow\sqrt{6t+7}=t\\ 6t+7-t^2=0\Leftrightarrow\left[\begin{array}{nghiempt}t=7\left(tm\right)\\t=-1\left(ktm\right)\end{array}\right.\)
Với \(t=7\Leftrightarrow x^2-2x-7=0\Leftrightarrow x=1\pm2\sqrt{2}\left(tm\right)\)
Vậy S={\(1\pm2\sqrt{2}\)}
a) \(\frac{1}{2}\sqrt{x-1}-\frac{3}{2}\sqrt{9x-9}+24\sqrt{\frac{x-1}{64}}=-17\)
<=> \(\frac{1}{2}\sqrt{x-1}-\frac{3}{2}\sqrt{9\left(x-1\right)}+24\frac{\sqrt{x-1}}{\sqrt{64}}=-17\)
<=>\(\frac{1}{2}\sqrt{x-1}-\frac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)
<=>\(\sqrt{x-1}\left(\frac{1}{2}-\frac{9}{2}+\frac{6}{2}\right)=-17\)
<=>\(\sqrt{x-1}=-17\)
<=>x-1=17
<=>x=18
Vậy pt có nghiệm là x=18
\(a.ĐK:x-1\ge0\Leftrightarrow x\ge1\)
\(\frac{1}{2}\sqrt{x-1}-\frac{3}{2}\sqrt{9x-9}+24\sqrt{\frac{x-1}{64}}=-17\)
\(\Leftrightarrow\frac{1}{2}\sqrt{x-1}-\frac{27}{2}\sqrt{x-1}+24\sqrt{\frac{x-1}{64}}=-17\)
\(\Leftrightarrow\sqrt{x-1}\left(\frac{1}{2}-\frac{27}{2}+24\sqrt{\frac{1}{64}}\right)=-17\)
\(\Leftrightarrow\sqrt{x-1}.\left(-10\right)=-17\)
\(\Leftrightarrow\sqrt{x-1}=\frac{-17}{-10}=\frac{17}{10}\)
\(\Leftrightarrow x-1=\left(\frac{17}{10}\right)^2\)
\(\Leftrightarrow x=\frac{289}{100}+1=3,89\left(TM\right)\)
Vậy \(S=\left\{3,89\right\}\)
\(b.ĐK:x^2+2\ge0\)
\(\sqrt{9x^2+18}+2\sqrt{x^2+2}-\sqrt{25x^2+50}+3=0\)
\(\Leftrightarrow9\sqrt{x^2+2}+2\sqrt{x^2+2}-25\sqrt{x^2+2}=-3\)
\(\Leftrightarrow\sqrt{x^2+2}\left(9+2-25\right)=-3\)
\(\Leftrightarrow\sqrt{x^2+2}=\frac{-3}{-14}=\frac{3}{14}\)
\(\Leftrightarrow x^2+2=\left(\frac{3}{14}\right)^2\)
\(\Leftrightarrow x=\sqrt{\frac{9}{196}-2}=\sqrt{-\frac{383}{196}}\left(vl\right)\)
Vậy \(S=\varnothing\)
Mấy câu kia làm tương tự
a) \(\dfrac{3-\sqrt{x}}{x-9}=\dfrac{-\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=-\dfrac{1}{\sqrt{x+3}}\)(\(x\ge0,x\ne9\))
b) \(\dfrac{x-5\sqrt{x}+6}{\sqrt{x}-3}=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}{\sqrt{x}-3}=\sqrt{x}-2\left(x\ge0,x\ne9\right)\)
a) \(\dfrac{3-\sqrt{x}}{x-9}=\dfrac{3-\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=-\dfrac{1}{\sqrt{x}+3}\)
b) \(\dfrac{x-5\sqrt{x}+6}{\sqrt{x}-3}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\sqrt{x}-3}=\sqrt{x}-2\)
c) \(6-2x-\sqrt{9-6x+x^2}=6-2x-\sqrt{\left(3-x\right)^2}=6-2x-\left|3-x\right|\)
mà \(x< 3\Rightarrow3-x>0\Rightarrow6-2x-\left|3-x\right|=6-2x-3+x=3-x\)
ĐKXĐ:...
a. Đặt \(\left\{{}\begin{matrix}\sqrt{2x^2+4x+16}=a>0\\\sqrt{x+70}=b\ge0\end{matrix}\right.\)
\(\Rightarrow6x^2+10x-92=3a^2-2b^2\)
Pt trở thành:
\(3a^2-2b^2+ab=0\)
\(\Leftrightarrow\left(a+b\right)\left(3a-2b\right)=0\)
\(\Leftrightarrow3a=2b\)
\(\Leftrightarrow9\left(2x^2+4x+16\right)=4\left(x+70\right)\)
\(\Leftrightarrow...\)
b. ĐKXĐ: ...
Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\ge0\\\sqrt{1-x}=b\ge0\end{matrix}\right.\)
Phương trình trở thành:
\(a^2+2+ab=3a+b\)
\(\Leftrightarrow a^2-3a+2+ab-b=0\)
\(\Leftrightarrow\left(a-1\right)\left(a-2\right)+b\left(a-1\right)=0\)
\(\Leftrightarrow\left(a-1\right)\left(a+b-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=1\\a+b=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=1\\\sqrt{x+1}+\sqrt{1-x}=2\end{matrix}\right.\)
\(\Leftrightarrow...\)
1/PT (1) cho ta nhân tử x - y - 1:)
\(\left\{{}\begin{matrix}\left(17-3x\right)\sqrt{5-x}+\left(3y-14\right)\sqrt{4-y}=0\left(1\right)\\2\sqrt{2x+y+5}+3\sqrt{3x+2y+11}=x^2+6x+13\left(2\right)\end{matrix}\right.\)
ĐK: \(x\le5;y\le4\); \(2x+y+5\ge0;3x+2y+11\ge0\)
PT (1) \(\Leftrightarrow\left(17-3x\right)\left(\sqrt{5-x}-\sqrt{4-y}\right)-3\left(x-y-1\right)\sqrt{4-y}=0\)
\(\Leftrightarrow\left(3x-17\right)\left(\frac{x-y-1}{\sqrt{5-x}+\sqrt{4-y}}\right)-3\left(x-y-1\right)\sqrt{4-y}=0\)
\(\Leftrightarrow\left(x-y-1\right)\left(\frac{3x-17}{\sqrt{5-x}+\sqrt{4-y}}-3\sqrt{4-y}\right)=0\)
Dễ thấy cái ngoặc to < 0
Do đó x= y + 1
Thay xuống PT (2):\(y^2+8y+20=2\sqrt{3y+7}+3\sqrt{5y+14}\)\(\left(y+1\right)\left(y+2\right)=y^2+3y+2\)
ĐK: \(y\ge-\frac{7}{3}\) (để các căn thức được thỏa mãn)
PT (2) \(\Leftrightarrow y^2+3y+2+2\left(y+3-\sqrt{3y+7}\right)+3\left(y+4-\sqrt{5y+14}\right)=0\)
\(\Leftrightarrow\left(y^2+3y+2\right)\left(1+\frac{2}{y+3+\sqrt{3y+7}}+\frac{3}{y+4+\sqrt{5y+14}}\right)=0\)
Cái ngoặc to > 0 =>...
P/s: Is that true? Ko đúng thì chịu thua-_- Mất nửa tiếng đồng hồ để gõ bài này đấy:(
2/ĐK: \(x\ge-y;y\ge0\)
PT (1) \(\Leftrightarrow x\left(x+y\right)+\sqrt{x+y}=2y^2+\sqrt{2y}\)
\(\Leftrightarrow\left(x-y\right)\left(x+y\right)+y\left(x-y\right)+\sqrt{x+y}-\sqrt{2y}=0\)
\(\Leftrightarrow\left(x-y\right)\left(x+2y+\frac{1}{\sqrt{x+y}+\sqrt{2y}}\right)=0\)
Cái ngoặc to \(\ge y+\frac{1}{\sqrt{x+y}+\sqrt{2y}}>0\).
Do đó x = y \(\ge0\)
Thay xuống pt dưới: \(x^3-5x^2+14x-4=6\sqrt[3]{x^2-x+1}\)
Lập phương hai vế lên ra pt bậc 6, tuy nhiên cứ yên tâm, nghiệm rất đẹp: x = 1:)
Em đưa kết quả luôn: \(\left(x-1\right)\left(x^2-4x+7\right)\left(x^6-10x^5+56x^4-160x^3+272x^2-64x+40\right)=0\)
P/s: khúc cuối em ko còn cách nào khác nên đành lập phương:((