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\(\sqrt{x^2-3x+2}-\sqrt{x+3}=\sqrt{x-2}+\sqrt{x^2+2x-3}\)
\(\Leftrightarrow\left(\sqrt{x^2-3x+2}-\sqrt{x-2}\right)-\left(\sqrt{x^2+2x-3}+\sqrt{x+3}\right)=0\)
\(\Leftrightarrow\dfrac{\left(x^2-3x+2\right)-\left(x-2\right)}{\sqrt{x^2-3x+2}+\sqrt{x-2}}-\dfrac{\left(x^2+2x-3\right)-\left(x+3\right)}{\sqrt{x^2+2x-3}-\sqrt{x+3}}=0\)
\(\Leftrightarrow\dfrac{\left(x-2\right)^2}{\sqrt{\left(x-2\right)\left(x-1\right)}+\sqrt{x-2}}-\dfrac{\left(x-2\right)\left(x+3\right)}{\sqrt{\left(x+3\right)\left(x-1\right)}-\sqrt{x+3}}=0\)
\(\Leftrightarrow\left(x-2\right)\left[\dfrac{x-2}{\sqrt{x-2}\left(\sqrt{x-1}+1\right)}-\dfrac{x+3}{\sqrt{x+3}\left(\sqrt{x-1}-1\right)}\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left[\dfrac{\sqrt{x-2}}{\sqrt{x-1}+1}-\dfrac{\sqrt{x+3}}{\sqrt{x-1}-1}\right]=0\)
Pt \(\dfrac{\sqrt{x-2}}{\sqrt{x-1}+1}-\dfrac{\sqrt{x+3}}{\sqrt{x-1}-1}=0\) vô no
(vì \(\dfrac{\sqrt{x-2}}{\sqrt{x-1}+1}< \dfrac{\sqrt{x+3}}{\sqrt{x-1}-1}\forall x\ge2\Rightarrow VT< 0\))
=> x - 2 = 0
<=> x = 2 (nhận)
\(\sqrt{4x+1}-\sqrt{3x-2}=\dfrac{x+3}{5}\)
\(\Leftrightarrow\dfrac{\left(4x+1\right)-\left(3x-2\right)}{\sqrt{4x+1}+\sqrt{3x-2}}-\dfrac{x+3}{5}=0\)
\(\Leftrightarrow\dfrac{x+3}{\sqrt{4x+1}+\sqrt{3x-2}}-\dfrac{x+3}{5}=0\)
\(\Leftrightarrow\left(\dfrac{1}{\sqrt{4x+1}+\sqrt{3x-2}}-\dfrac{1}{5}\right)\left(x+3\right)=0\)
TH1:
x + 3 = 0
<=> x = - 3 (loại)
TH2:
\(\dfrac{1}{\sqrt{4x+1}+\sqrt{3x-2}}-\dfrac{1}{5}=0\)
\(\Leftrightarrow\sqrt{4x+1}+\sqrt{3x-2}=5\)
\(\Leftrightarrow\left(\sqrt{4x+1}-3\right)+\left(\sqrt{3x-2}-2\right)=0\)
\(\Leftrightarrow\dfrac{4x+1-9}{\sqrt{4x+1}+3}+\dfrac{3x-2-4}{\sqrt{3x-2}+2}=0\)
\(\Leftrightarrow\dfrac{4\left(x-2\right)}{\sqrt{4x+1}+3}+\dfrac{3\left(x-2\right)}{\sqrt{3x-2}+2}=0\)
\(\Leftrightarrow\left(\dfrac{4}{\sqrt{4x+1}+3}+\dfrac{3}{\sqrt{3x-2}+2}\right)\left(x-2\right)=0\)
Pt \(\dfrac{4}{\sqrt{4x+1}+3}+\dfrac{3}{\sqrt{3x-2}+2}>0\forall x\ge\dfrac{2}{3}\) => vô no
=> x - 2 = 0
<=> x = 2 (nhận)
~ ~ ~
Vậy x = 2
Câu 1:
\(\Leftrightarrow\sqrt{2\left(x+1\right)\left(x+3\right)}+\sqrt{\left(x-1\right)\left(x+1\right)}=2\left(x+1\right)\)
- Với \(x< -1\Rightarrow\left\{{}\begin{matrix}VT\ge0\\VP< 0\end{matrix}\right.\) pt vô nghiệm
- Nhận thấy \(x=-1\) là 1 nghiệm
- Nếu \(x>-1\) kết hợp ĐKXĐ các căn thức ta được \(x\ge1\), pt tương đương:
\(\sqrt{2\left(x+3\right)}+\sqrt{x-1}=2\sqrt{x+1}\)
\(\Leftrightarrow2x+6+x-1+2\sqrt{2\left(x+3\right)\left(x-1\right)}=4x+4\)
\(\Leftrightarrow2\sqrt{2x^2+4x-6}=x-1\)
\(\Leftrightarrow4\left(2x^2+4x-6\right)=\left(x-1\right)^2\)
\(\Leftrightarrow7x^2+18x-25=0\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=-\frac{25}{7}< 0\left(l\right)\end{matrix}\right.\)
Vậy pt có nghiệm \(x=\pm1\)
Câu 2:
ĐKXĐ: \(x\ge1\)
\(\sqrt{x-1+2\sqrt{x-1}+1}-\sqrt{x-1-2\sqrt{x-1}+1}=2\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}+1\right)^2}-\sqrt{\left(\sqrt{x-1}-1\right)^2}=2\)
\(\Leftrightarrow\sqrt{x-1}+1-\left|\sqrt{x-1}-1\right|=2\)
- Nếu \(\sqrt{x-1}-1\ge0\Leftrightarrow x\ge2\) pt trở thành:
\(\sqrt{x-1}+1-\sqrt{x-1}+1=2\Leftrightarrow2=2\) (luôn đúng)
- Nếu \(1\le x< 2\) pt trở thành:
\(\sqrt{x-1}+1-1+\sqrt{x-1}=2\Leftrightarrow x=2\left(l\right)\)
Vậy nghiệm của pt là \(x\ge2\)
Câu 3:
Bình phương 2 vế ta được:
\(2x^2+2x+5+2\sqrt{\left(x^2+x+4\right)\left(x^2+x+1\right)}=2x^2+2x+9\)
\(\Leftrightarrow\sqrt{\left(x^2+x+4\right)\left(x^2+x+1\right)}=2\)
\(\Leftrightarrow\left(x^2+x+4\right)\left(x^2+x+1\right)=4\)
Đặt \(x^2+x+1=a>0\) pt trở thành:
\(a\left(a+3\right)=4\Leftrightarrow a^2+3a-4=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-4\left(l\right)\end{matrix}\right.\)
\(\Rightarrow x^2+x+1=1\Leftrightarrow x^2+x=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
Câu 5:
ĐKXĐ: \(x\ge1\)
\(\sqrt{x-1-4\sqrt{x-1}+4}+\sqrt{x-1-6\sqrt{x-1}+9}=1\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}-2\right)^2}+\sqrt{\left(\sqrt{x-1}-3\right)^2}=1\)
\(\Leftrightarrow\left|\sqrt{x-1}-2\right|+\left|\sqrt{x-1}-3\right|=1\)
Mà \(VT=\left|\sqrt{x-1}-2\right|+\left|3-\sqrt{x-1}\right|\ge\left|\sqrt{x-1}-2+3-\sqrt{x-1}\right|=1\)
\(\Rightarrow VT\ge VP\Rightarrow\) Đẳng thức xảy ra khi và chỉ khi:
\(\left\{{}\begin{matrix}\sqrt{x-1}-2\ge0\\\sqrt{x-1}-3\le0\end{matrix}\right.\) \(\Rightarrow5\le x\le10\)
Vậy nghiệm của pt là \(5\le x\le10\)
\(A=\sqrt{2x^2-4x+3}+3\)
Ta có: \(2x^2-4x+3\)
\(=2\left(x^2-2x+\frac{3}{2}\right)\)
\(=2\left(x^2-2.x.1+1^2+\frac{1}{2}\right)\)
\(=2[\left(x-1\right)^2+\frac{1}{2}]\)
\(=2\left(x-1\right)^2+1\ge1\)
\(\Rightarrow\sqrt{2\left(x-1\right)^2+1}\ge\sqrt{1}\)
\(\Rightarrow\sqrt{2\left(x-1\right)^2+1}+3\ge3+\sqrt{1}=4\)
\(\Rightarrow MinA=4\Leftrightarrow x=1\)
\(b.\sqrt[3]{x-17}+\sqrt{x+8}=5\) \(\left(ĐK:x\ge-8\right)\)
Đặt: \(a=\sqrt[3]{x-17},b=\sqrt{x+8}\)
\(\Rightarrow x-17=a^3,x+8=b^2\)
Khi đó:
\(\left\{{}\begin{matrix}a+b=5\\a^3-b^2=x-17-x-8=-25\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=5-b\\a^3-b^2=-25\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=5-b\\\left(5-b\right)^3-b^2=-25\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=5-b\\b^3-14b^2+75b-150=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=5-b\\b^3-5b^2-9b^2+45b+30b-150=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=5-b\\b^2\left(b-5\right)-9b\left(b-5\right)+30\left(b-5\right)=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=5-b\\\left(b-5\right)\left(b^2-9b+30\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=5-b\\\left[{}\begin{matrix}b=5\\b^2-9b+30=\left(b-\dfrac{9}{2}\right)^2+\dfrac{39}{4}=0\left(l\right)\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0\\b=5\end{matrix}\right.\)
Thế vào ta được:
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt[3]{x-17}=0\\\sqrt{x+8}=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x-17=0\\x+8=25\end{matrix}\right.\) \(\Leftrightarrow x=17\left(n\right)\)
a, ĐKXĐ : \(x\ge\dfrac{1}{2}\)
PT <=> 2x - 1 = 5
<=> x = 3 ( TM )
Vậy ...
b, ĐKXĐ : \(x\ge5\)
PT <=> x - 5 = 9
<=> x = 14 ( TM )
Vậy ...
c, PT <=> \(\left|2x+1\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)
Vậy ...
d, PT<=> \(\left|x-3\right|=3-x\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=x-3\\x-3=3-x\end{matrix}\right.\)
Vậy phương trình có vô số nghiệm với mọi x \(x\le3\)
e, ĐKXĐ : \(-\dfrac{5}{2}\le x\le1\)
PT <=> 2x + 5 = 1 - x
<=> 3x = -4
<=> \(x=-\dfrac{4}{3}\left(TM\right)\)
Vậy ...
f ĐKXĐ : \(\left[{}\begin{matrix}x\le0\\1\le x\le3\end{matrix}\right.\)
PT <=> \(x^2-x=3-x\)
\(\Leftrightarrow x=\pm\sqrt{3}\) ( TM )
Vậy ...
a) \(\sqrt{2x-1}=\sqrt{5}\) (x \(\ge\dfrac{1}{2}\))
<=> 2x - 1 = 5
<=> x = 3 (tmđk)
Vậy S = \(\left\{3\right\}\)
b) \(\sqrt{x-5}=3\) (x\(\ge5\))
<=> x - 5 = 9
<=> x = 4 (ko tmđk)
Vậy x \(\in\varnothing\)
c) \(\sqrt{4x^2+4x+1}=6\) (x \(\in R\))
<=> \(\sqrt{\left(2x+1\right)^2}=6\)
<=> |2x + 1| = 6
<=> \(\left[{}\begin{matrix}\text{2x + 1=6}\\\text{2x + 1}=-6\end{matrix}\right.< =>\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{-7}{2}\end{matrix}\right.\)(tmđk)
Vậy S = \(\left\{\dfrac{5}{2};\dfrac{-7}{2}\right\}\)
c) \(\sqrt{\left(x-2\right)^2}=10\)
\(x-2=10\)
\(x=12\)
d) \(\sqrt{9x^2-6x+1}=15\)
\(\sqrt{\left(3x\right)^2-2.3x.1+1^2}=15\)
\(\sqrt{\left(3x-1\right)^2}=15\)
\(3x-1=15\)
\(3x=16\)
\(x=\dfrac{16}{3}\)
a) \(đk:x\ge0\)
\(pt\Leftrightarrow3\sqrt{2x}+4\sqrt{2x}-3\sqrt{2x}=12\)
\(\Leftrightarrow4\sqrt{2x}=12\Leftrightarrow\sqrt{2x}=3\Leftrightarrow2x=9\Leftrightarrow x=\dfrac{9}{2}\left(tm\right)\)
b) \(đk:x\ge-2\)
\(pt\Leftrightarrow3\sqrt{x+2}+12\sqrt{x+2}-2\sqrt{x+2}=26\)
\(\Leftrightarrow13\sqrt{x+2}=26\)
\(\Leftrightarrow\sqrt{x+2}=2\Leftrightarrow x+2=4\Leftrightarrow x=2\left(tm\right)\)
c) \(pt\Leftrightarrow\left|x-2\right|=10\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=10\\x-2=-10\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-8\end{matrix}\right.\)
d) \(pt\Leftrightarrow\sqrt{\left(3x-1\right)^2}=15\)
\(\Leftrightarrow\left|3x-1\right|=15\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=15\\3x-1=-15\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{16}{3}\\x=-\dfrac{14}{3}\end{matrix}\right.\)
e) \(đk:x\ge\dfrac{8}{3}\)
\(pt\Leftrightarrow3x+4=9x^2-48x+64\)
\(\Leftrightarrow9x^2-51x+60=0\)
\(\Leftrightarrow3\left(x-4\right)\left(5x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=\dfrac{5}{3}\left(ktm\right)\end{matrix}\right.\)
Bài 1:
a) ĐKXĐ: \(x\geq \frac{-3}{2}\)
PT \(\Leftrightarrow x^2+4x+5-2\sqrt{2x+3}=0\)
\(\Leftrightarrow x^2+2x+1+(2x+3)-2\sqrt{2x+3}+1=0\)
\(\Leftrightarrow (x+1)^2+(\sqrt{2x+3}-1)^2=0\)
Vì $(x+1)^2\geq 0; (\sqrt{2x+3}-1)^2\geq 0$ với mọi $x\geq \frac{-3}{2}$ nên để tổng của chúng bằng $0$ thì $(x+1)^2=(\sqrt{2x+3}-1)^2=0$
$\Leftrightarrow x=-1$
Vậy $x=-1$
b) ĐKXĐ: \(x^2-4x-8\geq 0\)
PT \(\Leftrightarrow 2(x^2-4x-8)-3\sqrt{x^2-4x-8}=2\)
Đặt \(\sqrt{x^2-4x-8}=a(a\geq 0)\) thì PT trở thành:
\(2a^2-3a=2\)
\(\Leftrightarrow 2a^2-3a-2=0\Leftrightarrow (a-2)(2a+1)=0\)
\(\Rightarrow a=2\) (do $a\geq 0$)
\(\Leftrightarrow x^2-4x-8=4\)
\(\Leftrightarrow x^2-4x-12=0\Leftrightarrow \left[\begin{matrix} x=6\\ x=-2\end{matrix}\right.\) (đều thỏa mãn)
Bài 2:
\(199-2x-x^2=200-(x^2+2x+1)=200-(x+1)^2\leq 200, \forall x\in\mathbb{Z}\)
\(\Rightarrow 4y^2=2+\sqrt{199-2x-x^2}\leq 2+\sqrt{200}\)
\(\Leftrightarrow y^2\leq \frac{2+\sqrt{200}}{4}< 9\)
\(\Rightarrow -3< y< 3\). Mà $y$ nguyên nên $y\in\left\{-2;-1;0;1;2\right\}$
Thay từng giá trị của $y$ vào PT ban đầu ta tìm được các cặp $(x,y)$ sau:
$(x,y)=(1,\pm 2); (-3,\pm 2); (13,\pm 1); (-15,\pm 1)$