Phương trình chứa căn - Hỏi đáp

\(\left\{{}\begin{matrix}\dfrac{x+y+1}{x+y}-\dfrac{x-y-1}{x-y}=0\left(1\right)\left(đk:x\ne\pm y\right)\\x+2y=3\left(2\right)\end{matrix}\right.\)

pt (2) <=> x = 3 - 2y

Thế (2) vào (1) ta có:

\(\dfrac{4-y}{3-y}-\dfrac{4-3y}{3-3y}=0\Leftrightarrow\dfrac{3y^2-15y+12-3y^2+13y-12}{3y^2-12y+9}=0\)

<=> -2y= 0 <=> \(y=0\)\(\Rightarrow x=3\)

Vậy hệ có nghiệm (3;0)

ĐKXĐ: \(x\ge2\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x-2}=a\ge0\\\sqrt{x+1}=b>0\end{matrix}\right.\) ta được:

\(2a-3b=ab-6\Leftrightarrow ab-2a+3b-6=0\)

\(\Leftrightarrow a\left(b-2\right)+3\left(b-2\right)=0\Leftrightarrow\left(a+3\right)\left(b-2\right)=0\)

\(\Leftrightarrow b-2=0\) (do \(a+3>0\) \(\forall a\ge0\) )

\(\Leftrightarrow\sqrt{x+1}=2\Leftrightarrow x=3\)

Vậy pt có nghiệm duy nhất \(x=3\)

\(x^2+x-2x\sqrt{x}+\sqrt{x^3-1}=0\) ( đk: \(x\ge1\)

\(\left(x-\sqrt{x}\right)^2+\sqrt{x^3-1}=0\)

mà

\(\left(x-\sqrt{x}\right)^2\ge0;\sqrt{x^3-1}\ge0\)

\(\Rightarrow x=1\)

a) 5x +3=2x-8 <=>5x-2x=-8-3 <=>3x=-11 <=> x=\(\dfrac{-11}{3}\)

b)6x-3(x+2)=5x+3<=> (6-3-5)x-9=0 <=> x=\(\dfrac{-9}{2}\)

c) (3x-9)(5x+10)=0<=> \(\left[{}\begin{matrix}3x-9=0\\5x+10=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

d)8x(x+2)+16(x+2)=0<=>(x+2)(8x+16)=0<=>\(\left[{}\begin{matrix}x=-2\\x=-2\end{matrix}\right.\)

e)x² -12x+35=0 <=>\(\left[{}\begin{matrix}x=7\\x=5\end{matrix}\right.\)

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\(\sqrt{5x^2+14x+9}\) chơ

d) \(5\sqrt{x}+\dfrac{5}{2\sqrt{x}}=2x+\dfrac{1}{2x}+4\)

\(\Leftrightarrow2\left(x+\dfrac{1}{4x}\right)+4=5\left(\sqrt{x}+\dfrac{1}{2\sqrt{x}}\right)\)

\(\Leftrightarrow2\left[\left(\sqrt{x}+\dfrac{1}{2\sqrt{x}}\right)^2-1\right]-5\left(\sqrt{x}+\dfrac{1}{2\sqrt{x}}\right)+4=0\)

\(\Leftrightarrow2\left(\sqrt{x}+\dfrac{1}{2\sqrt{x}}\right)^2-5\left(\sqrt{x}+\dfrac{1}{2\sqrt{x}}\right)+2=0\)

Đặt \(\sqrt{x}+\dfrac{1}{2\sqrt{x}}=a\left(a\ge\sqrt{2}\right)\)

\(\Rightarrow2a^2-5a+2=0\)

\(\Leftrightarrow\left(a-2\right)\left(2a-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=2\left(\text{nhận}\right)\\a=\dfrac{1}{2}\left(\text{loại}\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{x}+\dfrac{1}{2\sqrt{x}}=2\)

\(\Leftrightarrow2x-4\sqrt{x}+1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=\dfrac{2+\sqrt{2}}{2}\\\sqrt{x}=\dfrac{2-\sqrt{2}}{2}\end{matrix}\right.\) (nhận)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3+2\sqrt{2}}{2}\\x=\dfrac{3-2\sqrt{2}}{2}\end{matrix}\right.\) (nhận)

a) \(\sqrt{x^2-3x+3}+\sqrt{x^2-3x+6}=3\)

Đặt \(\sqrt{x^2-3x+3}=a;\sqrt{x^2-3x+6}=b\left(a;b>0\right)\)

\(\Rightarrow\left\{{}\begin{matrix}a+b=3\\b^2-a^2=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a+b=3\\\left(b+a\right)\left(b-a\right)=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}b+a=3\\b-a=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}b=2\\a=1\end{matrix}\right.\) (nhận)

\(\Rightarrow\sqrt{x^2-3x+3}=1\)

\(\Leftrightarrow x^2-3x+3=1\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)

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

b) \(\sqrt{3-x+x^2}-\sqrt{2+x-x^2}=1\)

Đặt \(\sqrt{3-x+x^2}=a;\sqrt{2+x-x^2}=b\left(a;b>0\right)\)

\(\Rightarrow\left\{{}\begin{matrix}a-b=1\\a^2+b^2=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=b+1\\\left(b^2+2b+1\right)+b^2-5=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=b+1\\2\left(b-1\right)\left(b+2\right)=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a=2\\b=1\end{matrix}\right.\) (vì \(b+2>0\)) (nhận)

\(\Rightarrow\sqrt{2+x-x^2}=1\)

\(\Leftrightarrow2+x-x^2=1\)

\(\Leftrightarrow x^2-x-1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt{5}}{2}\\x=\dfrac{1-\sqrt{5}}{2}\end{matrix}\right.\) (nhận)

Điều kiện \(x\in R\)

Lập phương 2 vế phương trình đã cho ta được :

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

\(\Leftrightarrow\sqrt[3]{2x-1}\sqrt[3]{x-1}\left(\sqrt[3]{2x-1}+\sqrt[3]{x-1}\right)=1\)

mà \(\sqrt[3]{2x-1}+\sqrt[3]{x-1}=\sqrt[3]{3x+1}\) nên ta có :

\(\sqrt[3]{2x-1}\sqrt[3]{x-1}\sqrt[3]{3x+1}=1\)

\(\Leftrightarrow\left(2x-1\right)\left(x-1\right)\left(3x+1\right)=1\)

\(\Leftrightarrow x\in\left\{0;\frac{7}{6}\right\}\)

Thử lại ta thấy \(x=\frac{7}{6}\) là nghiệm duy nhất của phương trình đã cho 

1/ ĐKXĐ: \(x\ge1\)

\(\sqrt{x-1-4\sqrt{x-1}+4}+\sqrt{x-1-2.3\sqrt{x-1}+9}=1\)

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

\(\Leftrightarrow\left|\sqrt{x-1}-2\right|+\left|\sqrt{x-1}-3\right|=1\)

\(\Leftrightarrow\left|\sqrt{x-1}-2\right|+\left|3-\sqrt{x-1}\right|=1\)

Mà \(\left|\sqrt{x-1}-2\right|+\left|3-\sqrt{x-1}\right|\ge\left|\sqrt{x-1}-2+3-\sqrt{x-1}\right|=1\)

Dấu "=" xảy ra khi và chỉ khi \(\left\{{}\begin{matrix}\sqrt{x-1}\ge2\\\sqrt{x-1}\le3\end{matrix}\right.\) \(\Rightarrow5\le x\le10\)

Vậy phương trình nghiệm đúng với mọi \(x\in\left[5;10\right]\)

2/ ĐKXĐ: \(x\ge\dfrac{5}{2}\)

Nhân 2 vế với \(\sqrt{2}\) ta được:

\(\sqrt{2x+4+6\sqrt{2x-5}}+\sqrt{2x-4-2\sqrt{2x-5}}=4\)

\(\Leftrightarrow\sqrt{2x-5+2.3\sqrt{2x-5}+9}+\sqrt{2x-5-2\sqrt{2x-5}+1}=4\)

\(\Leftrightarrow\sqrt{\left(\sqrt{2x-5}+3\right)^2}+\sqrt{\left(\sqrt{2x-5}-1\right)^2}=4\)

\(\Leftrightarrow\sqrt{2x-5}+3+\left|\sqrt{2x-5}-1\right|=4\)

\(\Leftrightarrow\sqrt{2x-5}+\left|\sqrt{2x-5}-1\right|=1\)

TH1: \(\sqrt{2x-5}\ge1\Rightarrow\sqrt{2x-5}+\sqrt{2x-5}-1=1\)

\(\Leftrightarrow\sqrt{2x-5}=1\Rightarrow2x=6\Rightarrow x=3\)

TH2: \(\sqrt{2x-5}< 1\Rightarrow\sqrt{2x-5}+1-\sqrt{2x-5}=1\Leftrightarrow1=1\) (đúng với mọi \(\dfrac{5}{2}\le x< 3\))

Vậy nghiệm của phương trình là \(\dfrac{5}{2}\le x\le3\)