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a) đkxđ \(x\ge1\)
pt đã cho \(\Leftrightarrow\left(\sqrt{2x-1}-3\right)+\left(\sqrt{x-1}-2\right)=0\)
\(\Leftrightarrow\dfrac{2x-10}{\sqrt{2x-1}+3}+\dfrac{x-5}{\sqrt{x-1}+2}=0\)
\(\Leftrightarrow\left(x-5\right)\left(\dfrac{2}{\sqrt{2x-1}+3}+\dfrac{1}{\sqrt{x-1}+2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\left(nhận\right)\\\dfrac{2}{\sqrt{2x-1}+3}+\dfrac{1}{\sqrt{x-1}+3}=0\end{matrix}\right.\)
Hiển nhiên pt thứ 2 vô nghiệm vì \(VT>0\) với mọi \(x\ge1\). Do đó pt đã cho có nghiệm duy nhất là \(x=5\)
b) đkxđ: \(x\ge-3\)
Để ý rằng \(x^2+2x+7=\left(x^2+1\right)+\left(2x+6\right)=\left(x^2+1\right)+2\left(x+3\right)\) nên nếu ta đặt \(\sqrt{x^2+1}=u\left(u\ge1\right)\) và \(\sqrt{x+3}=v\left(v\ge0\right)\) thì pt đã chot rở thành:
\(u^2+2v^2=3uv\)
\(\Leftrightarrow\left(u-v\right)\left(u-2v\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}u=v\\u=2v\end{matrix}\right.\)
Nếu \(u=v\) thì \(\sqrt{x^2+1}=\sqrt{x+3}\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge-3\\x^2+1=x+3\end{matrix}\right.\)
Mà \(x^2+1=x+3\) \(\Leftrightarrow x^2-x-2=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-2\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\) (nhận)
Nếu \(u=2v\) thì \(\sqrt{x^2+1}=2\sqrt{x+3}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-3\\x^2+1=4x+12\end{matrix}\right.\)
mà \(x^2+1=4x+12\)\(\Leftrightarrow x^2-4x-11=0\)
\(\Leftrightarrow x=2\pm\sqrt{15}\) (nhận)
Vậy pt đã cho có tập nghiệm \(S=\left\{2;-1;2\pm\sqrt{15}\right\}\)
a) \(\sqrt{2x-1}+\sqrt{x-1}=5\) (ĐK: \(x\ge1\))
\(\Leftrightarrow\left(\sqrt{2x-1}+\sqrt{x-1}\right)^2=5^2\)
\(\Leftrightarrow2x-1+x-1+2\sqrt{\left(2x-1\right)\left(x-1\right)}=25\)
\(\Leftrightarrow3x-2+2\sqrt{\left(2x-1\right)\left(x-1\right)}=25\)
\(\Leftrightarrow\sqrt{\left(2x-1\right)\left(x-1\right)}=\dfrac{27-3x}{2}\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{27-3x}{2}\ge0\\\left(2x-1\right)\left(x-1\right)=\left(\dfrac{27-3x}{2}\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}27-3x\ge0\\2x^2-2x-x+1=\dfrac{729-162x+9x^2}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x\le27\\8x^2-12x+4=9x^2-162x+729\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le9\\x^2-150x+725=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le9\\\left[{}\begin{matrix}x-5=0\\x-145=0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le9\\\left[{}\begin{matrix}x=5\left(tm\right)\\x=145\left(ktm\right)\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow x=5\)
\(a,4\left(x-3\right)^2-\left(2x-1\right)^2\ge12\)
\(\Leftrightarrow4x^2-24x+36-4x^2-4x+1\ge12\)
\(\Leftrightarrow-28x+37\ge12\)
\(\Leftrightarrow-28x\ge12-37\)
\(\Leftrightarrow-28x\ge-25\)
\(\Leftrightarrow x\le\dfrac{25}{28}\)
Vậy \(S=\left\{x\left|x\le\dfrac{25}{28}\right|\right\}\)
b, \(\left(x-4\right)\left(x+4\right)\ge\left(x+3\right)^2+5\)
\(\Leftrightarrow x^2-16\ge x^2+6x+9+5\)
\(\Leftrightarrow x^2-x^2-6x\ge9+5+16\)
\(\Leftrightarrow-6x\ge30\)
\(\Leftrightarrow x\le-5\)
Vậy \(S=\left\{x\left|x\le-5\right|\right\}\)
\(c,\left(3x-1\right)^2-9\left(x+2\right)\left(x-2\right)< 5x\)
\(\Leftrightarrow9x^2-6x-1-9x^2+36< 5x\)
\(\Leftrightarrow9x^2-9x^2-6x-5x+36+1< 0\)
\(\Leftrightarrow-11x+37< 0\)
\(\Leftrightarrow-11x< -37\)
\(\Leftrightarrow x>\dfrac{37}{11}\)
vậy \(S=\left\{x\left|x>\dfrac{37}{11}\right|\right\}\)
a: ĐKXĐ: y<=1/2
\(\left\{{}\begin{matrix}3\left(x-1\right)-\sqrt{1-2y}=1\\\left(x-1\right)+2\sqrt{1-2y}=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}6\left(x-1\right)-2\sqrt{1-2y}=2\\\left(x-1\right)+2\sqrt{1-2y}=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}7\left(x-1\right)=7\\\left(x-1\right)+2\sqrt{1-2y}=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x-1=1\\2\sqrt{1-2y}=5-1=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=2\\\sqrt{1-2y}=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=2\\1-2y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-\dfrac{3}{2}\left(nhận\right)\end{matrix}\right.\)
b:
ĐKXĐ: \(x\in R\)
\(\left\{{}\begin{matrix}\sqrt{x^2-2x+1}-3y=7\\2\left|x-1\right|-8y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\sqrt{\left(x-1\right)^2}-3y=7\\2\left|x-1\right|-8y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left|x-1\right|-3y=7\\2\left|x-1\right|-8y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2\left|x-1\right|-6y=14\\2\left|x-1\right|-8y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2y=13\\\left|x-1\right|-3y=7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{13}{2}\\\left|x-1\right|=3y+7=3\cdot\dfrac{13}{2}+7=\dfrac{39}{2}+7=\dfrac{53}{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{13}{2}\\x-1\in\left\{\dfrac{53}{2};-\dfrac{53}{2}\right\}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{13}{2}\\x\in\left\{\dfrac{55}{2};-\dfrac{51}{2}\right\}\end{matrix}\right.\)
c: ĐKXĐ: y>=4
\(\left\{{}\begin{matrix}2\left(x^2-x\right)+\sqrt{y-4}=0\\3\left(x^2-x\right)-2\sqrt{y-4}=-7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}4\left(x^2-x\right)+2\sqrt{y-4}=0\\3\left(x^2-x\right)-2\sqrt{y-4}=-7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}7\left(x^2-x\right)=-7\\2\left(x^2-x\right)+\sqrt{y-4}=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x^2-x=-1\\\sqrt{y-4}=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x^2-x+1=0\\y-4=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\left(vôlý\right)\\y=8\end{matrix}\right.\)
=>\(\left(x,y\right)\in\varnothing\)
lời giải
a)
\(\left(x+1\right)\left(2x-1\right)+x\le2x^2+3\)
\(\Leftrightarrow2x^2+x-1+x\le2x^2+3\)
\(\Leftrightarrow2x\le4\Rightarrow x\le2\)
\(\)b) \(\left(x+1\right)\left(x+2\right)\left(x+3\right)-x>x^3+6x^2-5\)
\(\left(x^2+3x+2\right)\left(x+3\right)-x>x^3+6x^2-5\)
\(x^3+3x^2+3x^2+9x+2x+6-x>x^3+6x^2-5\)
\(10x+6>-5\Rightarrow x>-\dfrac{11}{10}\)
c)Đkxđ: x≥0x\ge0
x+√x>(2√x+3)(√x−1)x+\sqrt{x}>\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)
⇔x+√x>2x+√x−3\Leftrightarrow x+\sqrt{x}>2x+\sqrt{x}-3
⇔x−3>0\Leftrightarrow x-3>0
⇔x>3\Leftrightarrow x>3. (tmđk).