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\(1)\) ĐKXĐ : \(x\ge3\)
\(\sqrt{x^2-4x+3}+\sqrt{x-1}=0\)
\(\Leftrightarrow\)\(\sqrt{\left(x^2-4x+4\right)-1}+\sqrt{x-1}=0\)
\(\Leftrightarrow\)\(\sqrt{\left(x-2\right)^2-1}+\sqrt{x-1}=0\)
\(\Leftrightarrow\)\(\sqrt{\left(x-2-1\right)\left(x-2+1\right)}+\sqrt{x-1}=0\)
\(\Leftrightarrow\)\(\sqrt{\left(x-3\right)\left(x-1\right)}+\sqrt{x-1}=0\)
\(\Leftrightarrow\)\(\sqrt{x-1}\left(\sqrt{x-3}+1\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}\sqrt{x-1}=0\\\sqrt{x-3}+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x\in\left\{\varnothing\right\}\end{cases}}}\)
Vậy \(x=1\)
\(2)\)\(\sqrt{x^2-2x+1}-\sqrt{x^2-6x+9}=10\)
\(\Leftrightarrow\)\(\sqrt{\left(x-1\right)^2}-\sqrt{\left(x-3\right)^2}=10\)
\(\Leftrightarrow\)\(\left|x-1\right|-\left|x-3\right|=10\)
+) Với \(\hept{\begin{cases}x-1\ge0\\x-3\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge1\\x\ge3\end{cases}\Leftrightarrow}x\ge3}\) ta có :
\(x-1-x+3=10\)
\(\Leftrightarrow\)\(0=8\) ( loại )
+) Với \(\hept{\begin{cases}x-1< 0\\x-3< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x< 1\\x< 3\end{cases}\Leftrightarrow}x< 1}\) ta có :
\(1-x+x-3=10\)
\(\Leftrightarrow\)\(0=12\) ( loại )
Vậy không có x thỏa mãn đề bài
Chúc bạn học tốt ~
PS : mới lp 8 sai đừng chửi nhé :v
a:
\(\Leftrightarrow\sqrt{\left(x-3\right)^2}=3\)
=>|x-3|=3
=>x-3=3 hoặc x-3=-3
=>x=0 hoặc x=6
b: \(\Leftrightarrow\sqrt{x-1+2\sqrt{x-1}+1}=2\)
=>\(\sqrt{\left(\sqrt{x-1}+1\right)^2}=2\)
=>\(\left|\sqrt{x-1}+1\right|=2\)
=>\(\left[{}\begin{matrix}\sqrt{x-1}+1=2\\\sqrt{x-1}+1=-2\left(loại\right)\end{matrix}\right.\Leftrightarrow\sqrt{x-1}=1\)
=>x-1=1
=>x=2
c:
ĐKXĐ: x>4/5
PT \(\Leftrightarrow\sqrt{\dfrac{5x-4}{x+2}}=2\)
=>\(\dfrac{5x-4}{x+2}=4\)
=>5x-4=4x+8
=>x=12(nhận)
d: ĐKXĐ: x-4>=0 và x+1>=0
=>x>=4
PT =>\(\left(\sqrt{x-4}+\sqrt{x+1}\right)^2=5^2=25\)
=>\(x-4+x+1+2\sqrt{\left(x-4\right)\left(x+1\right)}=25\)
=>\(\sqrt{4\left(x^2-3x-4\right)}=25-2x+3=28-2x\)
=>\(\sqrt{x^2-3x-4}=14-x\)
=>x<=14 và x^2-3x-4=(14-x)^2=x^2-28x+196
=>x<=14 và -3x-4=-28x+196
=>x<=14 và 25x=200
=>x=8(nhận)
a) \(\sqrt{x^2-6x+9}=3\)
\(\Leftrightarrow\sqrt{\left(x-3\right)^2}=3\)
\(\Leftrightarrow\left|x-3\right|=3 \)
TH1: \(\left|x-3\right|=x-3\) với \(x\ge3\)
Pt trở thành:
\(x-3=3\) (ĐK: \(x\ge3\))
\(\Leftrightarrow x=3+3\)
\(\Leftrightarrow x=6\left(tm\right)\)
TH2: \(\left|x-3\right|=-\left(x-3\right)\) với \(x< 3\)
Pt trở thành:
\(-\left(x-3\right)=3\) (ĐK: \(x< 3\))
\(\Leftrightarrow x-3=-3\)
\(\Leftrightarrow x=-3+3\)
\(\Leftrightarrow x=0\left(tm\right)\)
b) \(\sqrt{x+2\sqrt{x-1}}=2\) (ĐK: \(x\ge1\))
\(\Leftrightarrow x+2\sqrt{x-1}=4\)
\(\Leftrightarrow2\sqrt{x-1}=4-x\)
\(\Leftrightarrow4\left(x-1\right)=16-8x+x^2\)
\(\Leftrightarrow4x-4=16-8x+x^2\)
\(\Leftrightarrow x^2-12x+20=0\)
\(\Leftrightarrow\left(x-10\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=10\left(tm\right)\\x=2\left(tm\right)\end{matrix}\right.\)
c) \(\dfrac{\sqrt{5x-4}}{\sqrt{x+2}}=2\) (ĐK: \(x\ge\dfrac{4}{5}\))
\(\Leftrightarrow\dfrac{5x-4}{x+2}=4\)
\(\Leftrightarrow5x-4=4x+8\)
\(\Leftrightarrow x=12\left(tm\right)\)
\(\sqrt{x^2-6x+9}-\sqrt{x^2-2x+1}=\sqrt{x^2}\)
\(\Rightarrow\sqrt{\left(x-3\right)^2}-\sqrt{\left(x-1\right)^2}=x\)
\(\Rightarrow x-3-x+1-x=0\)
\(\Rightarrow-x=2\Rightarrow x=-2\)
Vậy......
\(pt\Leftrightarrow\sqrt{\left(x-3\right)^2}-\sqrt{\left(x-1\right)^2}=\sqrt{x^2}\)
\(\Leftrightarrow\left|x-3\right|-\left|x-1\right|-\left|x\right|=0\)
Xét \(x< 0\Leftrightarrow3-x+x-1+x=0\)
\(\Leftrightarrow x=-2\)(tm)
Xét \(0\le x< 1\)\(\Leftrightarrow3-x+x-1-x=0\)
\(\Leftrightarrow x=1\left(l\right)\)
Xét \(1< x\le3\Leftrightarrow3-x-x+1-x=0\)
\(\Leftrightarrow4=3x\Leftrightarrow x=\frac{4}{3}\)(tm)
Xét \(x\ge3\Leftrightarrow x-3-x+1-x=0\)
\(\Leftrightarrow x=-1\left(l\right)\)
a) Ta có: \(\sqrt{49\left(x^2-2x+1\right)}-35=0\)
\(\Leftrightarrow7\left|x-1\right|=35\)
\(\Leftrightarrow\left|x-1\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)
b)
ĐKXĐ: \(\left[{}\begin{matrix}x\ge3\\x\le-3\end{matrix}\right.\)
Ta có: \(\sqrt{x^2-9}-5\sqrt{x+3}=0\)
\(\Leftrightarrow\sqrt{x+3}\left(\sqrt{x-3}-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+3}=0\\\sqrt{x-3}=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-3=25\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(nhận\right)\\x=28\left(nhận\right)\end{matrix}\right.\)
c) ĐKXĐ: \(x\ge0\)
Ta có: \(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-1}{\sqrt{x}+3}\)
\(\Leftrightarrow x-1=x+\sqrt{x}-6\)
\(\Leftrightarrow\sqrt{x}-6=-1\)
\(\Leftrightarrow\sqrt{x}=5\)
hay x=25(nhận)
\(a,\sqrt{x-2\sqrt{x}-1}-\sqrt{x-1}=1.\)
\(\Rightarrow\sqrt{\left(\sqrt{x}-1\right)^2}-\sqrt{x-1}=1\)
\(\Rightarrow x-1-\sqrt{x-1}=1\)
\(\Rightarrow\sqrt{x-1}=x-1+1\)
\(\Rightarrow x-1=x^2\Rightarrow x^2-x+1=0\) ( vô nghiệm vì nó luôn lớn hơn 0 )
\(đkxđ\Leftrightarrow2x-1\ge0\Rightarrow x\ge\frac{1}{2}\)
\(c,\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=\sqrt{2}.\)
\(\Rightarrow\sqrt{2x+2\sqrt{2x-1}}+\sqrt{2x-2\sqrt{2x-1}}=2\)
\(\Rightarrow\sqrt{2x-1+2\sqrt{2x-1}+1}+\sqrt{2x-1-2\sqrt{2x-1}+1}=2\)
\(\Rightarrow\sqrt{\left(\sqrt{2x-1}+1\right)^2}+\sqrt{\left(\sqrt{2x-1}-1\right)^2}=2\)
\(\Rightarrow\sqrt{2x-1}+1+\sqrt{2x-1}-1=2\)
\(\Rightarrow\sqrt{2x-1}+\sqrt{2x-1}=2\)
\(\Rightarrow\sqrt{2x-1}=1\Rightarrow\sqrt{2x-1}^2=1\)
\(\Rightarrow2x-1=1\Rightarrow2x=2\Leftrightarrow x=1\)\(\left(tm\right)\)
d tương tự nha , nhân thêm 2 vế với \(\sqrt{6}\)là ra
\(\sqrt{x^2+2x+5}=-x^2-2x+1\)
\(\Leftrightarrow\sqrt{\left(x+1\right)^2+4}=-\left(x+1\right)^2+2\)
Ta thấy :
\(-\left(x+1\right)^2+2\le2\) Với \(\forall x\in R\)
\(\sqrt{\left(x+1\right)^2+4}\ge2\) Với \(\forall x\in R\)
\(\Rightarrow\sqrt{\left(x+1\right)^2+4}=-\left(x+1\right)^2+2\) Khi x + 1 = 0 \(\Leftrightarrow\) x = -1
Vậy Phương trình có nghiệm x = -1 .
\(\sqrt{x^2-6x+10}+\sqrt{4x^2-24x+45}=-x^2+6x-5\)
Ta thấy :
\(\sqrt{x^2-6x+10}=\sqrt{\left(x-3\right)^2+1}\) \(\ge1\) Với \(\forall x\in R\)
\(\sqrt{4x^2-24x+45}=\sqrt{4\left(x-3\right)^2+9}\ge3\) Với \(\forall x\in R\)
\(-x^2+6x-5=-\left(x-3\right)^2+4\le4\) Với \(\forall x\in R\)
\(\Rightarrow VT\ge4\) ; \(VP\le4\)
\(\Rightarrow VT=VP=4\)
Dấu "=" xảy ra khi x - 3 = 0 \(\Leftrightarrow\) x = 3
Vậy phương trình có nghiệm x = 3 .
\(ĐK:x\in R\)
\(\sqrt{x^2+x+4}+\sqrt{x^2+x+1}=\sqrt{2x^2+2x+9}\) (*)
Đặt \(x^2+x+1=a;a\ge0\)
\(\rightarrow\left\{{}\begin{matrix}x^2+x+4=a+3\\2x^2+2x+9=2a+7\end{matrix}\right.\)
(*) \(\Rightarrow\sqrt{a+3}+\sqrt{a}=\sqrt{2a+7}\)
\(\Leftrightarrow\left(\sqrt{a+3}+\sqrt{a}\right)^2=\left(\sqrt{2a+7}\right)^2\)
\(\Leftrightarrow a+3+a+2\sqrt{a\left(a+3\right)}=2a+7\)
\(\Leftrightarrow2\sqrt{a\left(a+3\right)}=4\)
\(\Leftrightarrow\sqrt{a\left(a+3\right)}=2\)
\(\Leftrightarrow a\left(a+3\right)=4\)
\(\Leftrightarrow a^2+3a-4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=1\left(tm\right)\\a=-4\left(ktm\right)\end{matrix}\right.\)
\(\Rightarrow x^2+x+1=1\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\) \((tm)\)
Vậy \(S=\left\{0;-1\right\}\)
a/ ĐKXĐ: \(\left[{}\begin{matrix}x\ge-1\\x\le-5\end{matrix}\right.\)
Bình phương 2 vế:
\(x^2+3x+2+2\sqrt{\left(x^2+3x+2\right)\left(x^2+6x+5\right)}+x^2+6x+5=2x^2+9x+7\)
\(\Leftrightarrow2\sqrt{\left(x^2+3x+2\right)\left(x^2+6x+5\right)}=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2+3x+2=0\\x^2+6x+5=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-1\\x=-2\left(l\right)\\x=-5\end{matrix}\right.\)
Vậy pt có 2 nghiệm \(x=-1;x=-5\)
b/ ĐKXĐ: \(x\ge-1\)
Đặt \(\sqrt{2x+3}+\sqrt{x+1}=a>0\Rightarrow a^2-6=3x+2\sqrt{2x^2+5x+3}-2\)
Phương trình trở thành:
\(a=a^2-6\Leftrightarrow a^2-a-6=0\Rightarrow\left[{}\begin{matrix}a=-2\left(l\right)\\a=3\end{matrix}\right.\)
\(\Rightarrow\sqrt{2x+3}+\sqrt{x+1}=3\Leftrightarrow3x+4+2\sqrt{2x^2+5x+3}=9\)
\(\Leftrightarrow2\sqrt{2x^2+5x+3}=5-3x\)
\(\Leftrightarrow\left\{{}\begin{matrix}5-3x\ge0\\4\left(2x^2+5x+3\right)=\left(5-3x\right)^2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{5}{3}\\x^2-50x+13=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=25+6\sqrt{17}\left(l\right)\\x=25-6\sqrt{17}\end{matrix}\right.\)
Vậy pt có nghiệm duy nhất \(x=25-6\sqrt{17}\)
a) \(\sqrt{\left(x+1\right)\left(x+2\right)}+\sqrt{\left(x+1\right)\left(x+5\right)}=\sqrt{\left(x+1\right)\left(2x+7\right)}\)
\(ĐK\Leftrightarrow\left[{}\begin{matrix}x\le-1\\x\ge-2\end{matrix}\right.\)
\(\Leftrightarrow\sqrt{\left(x+1\right)\left(x+2\right)}+\sqrt{\left(x+1\right)\left(x+5\right)}-\sqrt{\left(x+1\right)\left(2x+7\right)}=0\)
\(\Leftrightarrow\sqrt{\left(x+1\right)}\left(\sqrt{x+2}+\sqrt{x+5}-\sqrt{2x+7}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\\sqrt{x+2}+\sqrt{x+5}=\sqrt{2x+7}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x+2+x+5+2\sqrt{\left(x+2\right)\left(x+5\right)}=2x+7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\2\sqrt{\left(x+2\right)\left(x+5\right)}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-2\\x=-5\end{matrix}\right.\)
vậy \(S=\left\{-1;-2;-5\right\}\)
đơn giản như đan rổ
1. đk: pt luôn xác định với mọi x
\(\sqrt{x^2-2x+1}-\sqrt{x^2-6x+9}=10\)
\(\Leftrightarrow\sqrt{\left(x-1\right)^2}-\sqrt{\left(x-3\right)^2}=10\)
\(\Leftrightarrow\left|x-1\right|-\left|x-3\right|=10\)
Bạn mở dấu giá trị tuyệt đối như lớp 7 là ok rồi!
2. đk: \(x\geq 1\)
\(\sqrt{x+2\sqrt{x-1}}=3\sqrt{x-1}-5\)
\(\Leftrightarrow\sqrt{x-1+2\sqrt{x-1}+1}=3\sqrt{x-1}-5\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}-1\right)^2}-3\sqrt{x-1}+5=0\)
\(\Leftrightarrow\left|\sqrt{x-1}-1\right|-3\sqrt{x-1}+5=0\)
Đến đây thì ổn rồi! bạn cứ xét khoảng rồi mở trị và bình phương 1 chút là ok cái bài!