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Câu a:
ĐKXĐ: \(x\geq 1\)
\(\sqrt{x-1}-\sqrt{5x-1}=\sqrt{3x-2}\)
\(\Leftrightarrow \sqrt{x-1}=\sqrt{3x-2}+\sqrt{5x-1}\)
\(\Rightarrow x-1=8x-3+2\sqrt{(3x-2)(5x-1)}\) (bình phương 2 vế)
\(\Leftrightarrow 7x-2+2\sqrt{(3x-2)(5x-1)}=0\)
(Vô lý với mọi \(x\geq 1\) )
Do đó PT vô nghiệm.
Câu b)
PT \(\Leftrightarrow \sqrt{3(x^2+2x+1)+4}+\sqrt{5(x^2+2x+1)+9}=5-(x^2+2x+1)\)
\(\Leftrightarrow \sqrt{3(x+1)^2+4}+\sqrt{5(x+1)^2+9}=5-(x+1)^2\)
Vì \((x+1)^2\geq 0, \forall x\) nên:
\(\sqrt{3(x+1)^2+4}\geq \sqrt{4}=2\)
\(\sqrt{5(x+1)^2+9}\geq \sqrt{9}=3\)
\(\Rightarrow \sqrt{3(x+1)^2+4}+\sqrt{5(x+1)^2+9}\geq 5(1)\)
Mặt khác ta cũng có: \(5-(x+1)^2\leq 5-0=5(2)\)
Từ \((1);(2)\Rightarrow \sqrt{3(x+1)^2+4}+\sqrt{5(x+1)^2+9}\geq 5\geq 5-(x+1)^2\)
Dấu "=" xảy ra khi $(x+1)^2=0$ hay $x=-1$ (thỏa mãn)
Vậy pt có nghiệm $x=-1$
Bài 1:
Để căn thức có nghĩa thì:
a)
\(-5x-10\geq 0\Leftrightarrow 5x+10\leq 0\Leftrightarrow x\leq -2\)
b)
\(x^2-3x+2\geq 0\Leftrightarrow (x-1)(x-2)\geq 0\)
\(\Leftrightarrow \left[\begin{matrix} x-1\geq 0; x-2\geq 0\\ x-1\leq 0; x-2\leq 0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x\geq 2\\ x\leq 1\end{matrix}\right.\)
c) \(\frac{x+3}{5-x}\geq 0\)
\(\Leftrightarrow \left[\begin{matrix} x+3\geq 0; 5-x>0\\ x+3\leq 0; 5-x< 0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} -3\leq x< 5\\ -3\geq x>5 (\text{vô lý})\end{matrix}\right.\)
\(\Rightarrow -3\leq x< 5\)
d) \(-x^2+4x-4\geq 0\)
\(\Leftrightarrow -(x^2-4x+4)\geq 0\Leftrightarrow -(x-2)^2\geq 0\)
Vì \((x-2)^2\geq 0, \forall x\in\mathbb{R}\)
\(\Rightarrow x=2\)
\(x^2-2-2\sqrt{4x-7}=0\)
\(\Leftrightarrow\left(4x-7-2\sqrt{4x-7}+1\right)+\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(\sqrt{4x-7}-1\right)^2+\left(x-2\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{4x-7}-1=0\\x-2=0\end{matrix}\right.\)
Tự làm tiếp nhé.
. . .
\(4x^2-5x+1+2\sqrt{x-1}=0\)
\(\Leftrightarrow\left(x-1\right)\left(4x-1\right)+2\sqrt{x-1}=0\)
\(\Leftrightarrow\sqrt{x-1}\left[\left(4x-1\right)\sqrt{x-1}+2\right]=0\)
\(\Rightarrow x=1\)
. . .
\(\sqrt{x^2-4x+4}+\sqrt{x^2-6x+9}=1\)
\(\Leftrightarrow\sqrt{\left(x-2\right)^2}+\sqrt{\left(x-3\right)^2}=1\)
\(\Leftrightarrow\left|x-2\right|+\left|x-3\right|=1\)
\(VT=\left|x-2\right|+\left|3-x\right|\ge\left|x-2+3-x\right|=1=VP\)
Dấu "=" xảy ra khi \(\left(x-2\right)\left(3-x\right)\ge0\)
Đến đây lập bảng xét dấu
. . .
\(x^2-x+2=2\sqrt{x^2-x+1}\)
\(\Leftrightarrow\left(\sqrt{x^2-x+1}-1\right)^2=0\)
Tự làm tiếp nhé.
\(\sqrt{3x+1}-\sqrt{6-x}+3x^2-14x-8=0\)
\(\Leftrightarrow\left(\sqrt{3x+1}-4\right)+\left(1-\sqrt{6-x}\right)+\left(3x^2-14-5\right)=0\)
\(\Leftrightarrow\dfrac{3x+1-16}{\sqrt{3x+1}+4}+\dfrac{1-6+x}{1+\sqrt{6-x}}+\left(x-5\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\dfrac{3\left(x-5\right)}{\sqrt{3x+1}+4}+\dfrac{x-5}{1+\sqrt{6-x}}+\left(x-5\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left(\dfrac{3}{\sqrt{3x+1}+4}+\dfrac{1}{1+\sqrt{6-x}}+3x+1\right)\left(x-5\right)=0\)
\(\Rightarrow x=5\)
. . .
\(\sqrt{2x^2-4x+5}-x+4=0\)
\(\Leftrightarrow\sqrt{2x^2-4x+5}=x-4\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-4\ge0\\2x^2-4x+5=x^2-8x+16\end{matrix}\right.\)
Tự làm tiếp nhé.
. . .
\(\sqrt{2x+3}+\sqrt{x-1}=\sqrt{x+6}\)
\(\Leftrightarrow\sqrt{2x+3}=\sqrt{x+6}-\sqrt{x-1}\)
\(\Leftrightarrow2x+3=x+6-2\sqrt{\left(x+6\right)\left(x-1\right)}+x-1\)
\(\Leftrightarrow2\sqrt{x^2+5x-6}=2\)
\(\Leftrightarrow x^2+5x-6=1\)
Tự làm tiếp nhé.
. . .
\(x+y+\dfrac{1}{2}=\sqrt{x}+\sqrt{y}\)
\(\Leftrightarrow\left(x-\sqrt{x}+\dfrac{1}{4}\right)+\left(y-\sqrt{y}+\dfrac{1}{4}\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\left(\sqrt{y}-\dfrac{1}{2}\right)^2=0\)
Tự làm tiếp nhé.
a) \(\sqrt{4x}=10\) (ĐKXĐ: 4x>=0 <=> x>=0)
\(\Leftrightarrow4x=100\)
\(\Leftrightarrow x=25\)
\(S=\left\{25\right\}\)
b) \(\sqrt{x^2-2x+1}=8\)
\(\Leftrightarrow\sqrt{\left(x-1\right)^2}=8\)
\(\Leftrightarrow x-1=8\)
\(\Leftrightarrow x=9\)
\(S=\left\{9\right\}\)
c) \(\sqrt{x^2-6x+9}=\sqrt{1-6x+9x^2}\)
\(\Leftrightarrow\sqrt{\left(x-3\right)^2}=\sqrt{\left(1-3x\right)^2}\)
\(\Leftrightarrow x-3=1-3x\) hoặc \(\Leftrightarrow x-3=-1+3x\)
\(\Leftrightarrow x+3x=1+3\) \(\Leftrightarrow x-3x=-1+3\)
\(\Leftrightarrow4x=4\) \(\Leftrightarrow-2x=2\)
\(\Leftrightarrow x=1\) \(\Leftrightarrow x=-1\)
\(S=\left\{1;-1\right\}\)
d) \(\sqrt{2x-5}=x-2\)
\(\Leftrightarrow2x-5=x^2-4x+4\)
\(\Leftrightarrow-x^2+2x+4x-5-4=0\)
\(\Leftrightarrow-x^2+6x-9=0\)
\(\Leftrightarrow x^2-6x+9=0\)
\(\Leftrightarrow\left(x-3\right)^2=0\)
\(\Leftrightarrow x-3=0\)
\(\Leftrightarrow x=3\)
\(S=\left\{3\right\}\)
e) \(\sqrt{x^2-2x+1}=\sqrt{x+1}\)
\(\Leftrightarrow x^2-2x+1=x+1\)
\(\Leftrightarrow x^2-2x-x+1-1=0\)
\(\Leftrightarrow x^2-3x=0\)
\(\Leftrightarrow x\left(x-3\right)=0\)
\(\Leftrightarrow x=0\) hoặc \(\Leftrightarrow x-3=0\)
\(\Leftrightarrow x=3\)
\(S=\left\{0;3\right\}\)
g) \(\sqrt{x^2-9}-\sqrt{x-3}=0\) ( ĐKXĐ: x-3>=0 <=> x>=3)
\(\Leftrightarrow\sqrt{x^2-9}=\sqrt{x-3}\)
\(\Leftrightarrow x^2-9=x-3\)
\(\Leftrightarrow x^2-x-6=0\)
\(\Leftrightarrow x^2-3x+2x-6=0\)
\(\Leftrightarrow\left(x^2+2x\right)-\left(3x+6\right)=0\)
\(\Leftrightarrow x\left(x+2\right)-3\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow x+2=0\) hoặc \(\Leftrightarrow x-3=0\)
\(\Leftrightarrow x=-2\) \(\Leftrightarrow x=3\)
\(S=\left\{-2;3\right\}\)
h) \(\sqrt{x^2-4x+4}+\sqrt{x^2-6x+9}=1\)
\(\Leftrightarrow\sqrt{\left(x-2\right)^2}+\sqrt{\left(x-3\right)^2}=1\)
\(\Leftrightarrow x-2+x-3-1=0\)
\(\Leftrightarrow2x-6=0\)
\(\Leftrightarrow x=3\)
\(S=\left\{3\right\}\)
i) \(\sqrt{\frac{2x-3}{x-1}}=2\)
\(\Leftrightarrow\frac{2x-3}{x-1}=4\)
\(\Leftrightarrow4\left(x-1\right)=2x-3\)
\(\Leftrightarrow4x-4-2x+3=0\)
\(\Leftrightarrow2x-1=0\)
\(\Leftrightarrow x=\frac{1}{2}\)
\(S=\left\{\frac{1}{2}\right\}\)
l) \(x+y+12=4\sqrt{x}+6\sqrt{y-1}\)
\(\Leftrightarrow x+y-4\sqrt{x}+12-6\sqrt{y-1}=0\)
\(\Leftrightarrow\left(x-4\sqrt{x}+4\right)+\left(y-1-6\sqrt{y-1}+9\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)^2+\left(\sqrt{y-1}-3\right)^2=0\)
\(\Leftrightarrow\sqrt{x}-2=0\) hoặc \(\Leftrightarrow\sqrt{y-1}-3=0\)
\(\Leftrightarrow\sqrt{x}=2\) \(\Leftrightarrow\sqrt{y-1}=3\)
\(\Leftrightarrow x=4\) \(\Leftrightarrow y-1=9\)
\(\Leftrightarrow y=10\)
KẾT luận : ..............
Tới đây nhé, nếu mai chưa ai giải thì mình giải hộ cho
CHÚC BẠN HỌC TỐT!
m) \(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8+6\sqrt{x-1}}=5\)
<=> \(\sqrt{\left(x-1\right)-4\sqrt{x-1}+4}+\sqrt{\left(x-1\right)+6\sqrt{x-1}+9}=5\)
<=>\(\sqrt{\left(\sqrt{x-1}+2\right)^2}+\sqrt{\left(\sqrt{x-1}+3\right)^2}=5\)
<=>\(\sqrt{x-1}+2+\sqrt{x-1}+3=5\)
<=> \(2\sqrt{x-1}=0\)
<=> \(\sqrt{x-1}=0\) <=>x=1
Vậy \(S=\left\{1\right\}\)
n) \(\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=\sqrt{2}\) (*) ( đk \(x\ge\frac{1}{2}\))
<=> \(\left(\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}\right)^2=2\)
<=> \(x+\sqrt{2x-1}+x-\sqrt{2x-1}+2\sqrt{x^2-2x+1}=2\)
<=> 2x+\(2\sqrt{\left(x-1\right)^2=2}\)
<=> x+\(\left|x-1\right|=2\)(1)
TH1: \(\frac{1}{2}\le x\le1\)
Từ (1) => x+1-x=2
<=> 1=2(vô lý)
TH2: x>1
Từ (1)=> x+x-1=2
<=> 2x=3<=> \(x=\frac{2}{3}\)(tm pt (*))
Vậy \(S=\left\{\frac{2}{3}\right\}\)
p) \(\sqrt{2x-1}+\sqrt{x-2}=\sqrt{x+1}\) (*) (đk :\(x\ge2\))
Đặt \(\left\{{}\begin{matrix}x-2=a\left(a\ge0\right)\\x+1=b\left(b\ge0\right)\end{matrix}\right.\) =>a+b=2x-1
Có \(\sqrt{a+b}+\sqrt{a}=\sqrt{b}\)
<=> \(\sqrt{a+b}=\sqrt{b}-\sqrt{a}\)
<=> \(a+b=b-2\sqrt{ab}+a\)
<=> 0=\(-2\sqrt{ab}\)
=> \(\left[{}\begin{matrix}a=0\\b=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\) => x=2 (vì x=-1 không thỏa mãn pt(*))
Vậy \(S=\left\{2\right\}\)
q) \(\sqrt{x-7}+\sqrt{9-x}=x^2-16x+66\)(*) (đk : \(7\le x\le9\))
Với a,b\(\ge0\) có: \(\sqrt{a}+\sqrt{b}\le2\sqrt{\frac{a+b}{2}}\)(tự cm nha) .Dấu "=" xảy ra <=> a=b
Áp dụng bđt trên có:
\(\sqrt{x-7}+\sqrt{9-x}\le2\sqrt{\frac{x-7+9-x}{2}}=2\sqrt{\frac{2}{2}}=2\) (1)
Có x2-16x+66=(x2-16x+64)+2=(x-8)2+2 \(\ge2\) với mọi x (2)
Từ (1),(2) .Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}x-7=9-x\\x-8=0\end{matrix}\right.\)<=>\(\left\{{}\begin{matrix}2x=16\\x=8\end{matrix}\right.\)<=>\(\left\{{}\begin{matrix}x=8\\x=8\end{matrix}\right.\)<=> x=8( tm pt (*))
Vậy \(S=\left\{8\right\}\)
a) giải pt ra ta được : x=-1
b) giải pt ra ta được : x=2
c)giải pt ra ta được : x vô ngiệm
d)giải pt ra ta được : x=vô ngiệm
~~~~~~~~~~~ai đi ngang qua nhớ để lại k ~~~~~~~~~~~~~
~~~~~~~~~~~~ Chúc bạn sớm kiếm được nhiều điểm hỏi đáp ~~~~~~~~~~~~~~~~~~~
\(\sqrt{3x^2+6x+12}+\sqrt{5x^2-10x^2+9}=\sqrt{3\left(x^2+2x+1\right)+9}+\sqrt{5\left(x^2-2x+1\right)+4}\)
\(\ge\sqrt{9}+\sqrt{4}=3+2=5\)
a/ ĐKXĐ: ...
\(\sqrt{x-7}-\frac{1}{2}+\sqrt{x-5}-\frac{3}{2}=0\)
\(\Leftrightarrow\frac{x-\frac{29}{4}}{\sqrt{x-7}+\frac{1}{2}}+\frac{x-\frac{29}{4}}{\sqrt{x-5}+\frac{3}{2}}=0\)
\(\Leftrightarrow\left(x-\frac{29}{4}\right)\left(\frac{1}{\sqrt{x-7}+\frac{1}{2}}+\frac{1}{\sqrt{x-5}+\frac{3}{2}}\right)=0\)
\(\Leftrightarrow x=\frac{29}{4}\)
b/ \(\Leftrightarrow\sqrt{x^2-6x+9}=3x+2\left(x\ge-\frac{2}{3}\right)\)
\(\Leftrightarrow x^2-6x+9=9x^2+12x+4\)
\(\Leftrightarrow8x^2-18x-5=0\Rightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=-\frac{1}{4}\end{matrix}\right.\)
c/
\(\sqrt{3\left(x+1\right)^2+9}+\sqrt{5x^2\left(x^2+2\right)+9}=5-2\left(x+1\right)^2\)
Do \(\left\{{}\begin{matrix}3\left(x+1\right)^2+9\ge9\\5x^2\left(x^2+2\right)\ge9\end{matrix}\right.\) \(\Rightarrow VT\ge\sqrt{9}+\sqrt{9}=6\)
\(VP=5-2\left(x+1\right)^2\le5< VP\)
Pt luôn vô nghiệm
tui lớp 4 nên ko bít