Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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) ĐKXĐ : \(x\ge-1\)
\(\sqrt{16x+16}-\sqrt{9x+9}=4\)\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}=4\)
\(\Leftrightarrow\sqrt{x+1}=4\Leftrightarrow x+1=16\Leftrightarrow x=15\)
b) ĐKXĐ : \(x\ge\frac{2}{3}\)
\(\sqrt{3x-2}-\sqrt{x+7}=1\Leftrightarrow3x-2+x+7-2\sqrt{3x-2}.\sqrt{x+7}=1\)
\(\Leftrightarrow4x+4-2\sqrt{3x^2+19x-14}=0\)\(\Leftrightarrow2x+2-\sqrt{3x^2+19x-14}=0\)
\(\Leftrightarrow2x+2=\sqrt{3x^2+19x-14}\Leftrightarrow\left(2x+2\right)^2=3x^2+19x-14\)
\(\Leftrightarrow4x^2+8x+4=3x^2+19x-14\Leftrightarrow x^2-11x+18=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=9\\x=2\end{cases}\left(tm\right)}\)
c) câu cuối bình phương tương tự câu b
Ai hack nick mình thì trả lại đi !!!
nick :
- Tên: Vô danh
- Đang học tại: Trường Tiểu học Số 1 Nà Nhạn
- Địa chỉ: Huyện Điện Biên - Điện Biên
- Điểm hỏi đáp: 112SP, 0GP
- Điểm hỏi đáp tuần này: 47SP, 0GP
- Thống kê hỏi đáp
Ai hack hộ mình rồi gửi cho mình nhé mình cảm ơn
Ai là bạn của mình chắn chắn biết nên vào phần bạn bè hỏi mình mới là chủ nick
Mong olm xem xét ko cho ai hack nick nhau nữa ạ! Xin chân thành cảm ơn !
LInk : https://olm.vn/thanhvien/lehoangngantoanhoc
Giải pt :
1
a. ĐKXĐ : \(x\ge4\)
Ta có :
\(\sqrt{x+3}-\sqrt{x-4}=1\\ \Leftrightarrow\sqrt{x+3}=1+\sqrt{x-4}\\ \Leftrightarrow x+3=x-3+2\sqrt{x-4}\\ \Leftrightarrow6=2\sqrt{x-4}\)
\(\Leftrightarrow3=\sqrt{x-4}\\ \Leftrightarrow x-4=9\)
\(\Leftrightarrow x=13\) (TM ĐKXĐ)
Vậy \(S=\left\{13\right\}\)
b.ĐKXĐ : \(-3\le x\le10\)
Ta có :
\(\sqrt{10-x}+\sqrt{x+3}=5\\ \Leftrightarrow13+2\sqrt{-x^2+7x+30}=25\\ \Leftrightarrow\sqrt{-x^2+7x+30}=6\\ \Leftrightarrow-x^2+7x+30=36\\ \Leftrightarrow-x^2+7x-6=0\\ \Leftrightarrow-x^2+x+6x-6=0\\ \Leftrightarrow-x\left(x-1\right)+6\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(6-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\left(TMĐKXĐ\right)\\x=6\left(TMĐKXĐ\right)\end{matrix}\right.\)
Vậy \(S=\left\{1;6\right\}\)
Câu a:
ĐKXĐ:...........
\(\sqrt{x^2-x+9}=2x+1\)
\(\Rightarrow \left\{\begin{matrix} 2x+1\geq 0\\ x^2-x+9=(2x+1)^2=4x^2+4x+1\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x\geq \frac{-1}{2}\\ 3x^2+5x-8=0\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x\geq \frac{-1}{2}\\ 3x(x-1)+8(x-1)=0\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x\geq \frac{-1}{2}\\ (x-1)(3x+8)=0\end{matrix}\right.\Rightarrow x=1\)
Vậy.....
Câu b:
ĐKXĐ:.........
Ta có: \(\sqrt{5x+7}-\sqrt{x+3}=\sqrt{3x+1}\)
\(\Rightarrow (\sqrt{5x+7}-\sqrt{x+3})^2=3x+1\)
\(\Leftrightarrow 5x+7+x+3-2\sqrt{(5x+7)(x+3)}=3x+1\)
\(\Leftrightarrow 3(x+3)=2\sqrt{(5x+7)(x+3)}\)
\(\Leftrightarrow \sqrt{x+3}(3\sqrt{x+3}-2\sqrt{5x+7})=0\)
Vì \(x\geq -\frac{7}{5}\Rightarrow \sqrt{x+3}>0\). Do đó:
\(3\sqrt{x+3}-2\sqrt{5x+7}=0\)
\(\Rightarrow 9(x+3)=4(5x+7)\)
\(\Rightarrow 11x=-1\Rightarrow x=\frac{-1}{11}\) (thỏa mãn)
Vậy..........
1) \(\sqrt{\text{x^2− 20x + 100 }}=10\)
<=> \(\sqrt{\left(x-10\right)^2}=10\)
<=> \(\left|x-10\right|=10\)
=> \(\left[{}\begin{matrix}x-10=10\\x-10=-10\end{matrix}\right.\)=> \(\left[{}\begin{matrix}x=10+10\\x=\left(-10\right)+10\end{matrix}\right.\)=>\(\left[{}\begin{matrix}x=20\\x=0\end{matrix}\right.\)
Vậy S = \(\left\{20;0\right\}\)
2) \(\sqrt{x +2\sqrt{x}+1}=6\)
<=> \(\sqrt{\left(\sqrt{x^2}+2.\sqrt{x}.1+1^2\right)}=6\)
<=> \(\sqrt{\left(\sqrt{x}+1\right)^2}=6\)
<=> \(\left|\sqrt{x}+1\right|=6\)
=> \(\left[{}\begin{matrix}\sqrt{x}+1=6\\\sqrt{x}+1=-6\end{matrix}\right.\)=>\(\left[{}\begin{matrix}\sqrt{x}=6-1=5\\\sqrt{x}=\left(-6\right)-1=-7\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=25\\x=-49\left(loai\right)\end{matrix}\right.\)
Vậy S = \(\left\{25\right\}\)
3) \(\sqrt{x^2-6x+9}=\sqrt{4+2\sqrt{3}}\)
<=> \(\sqrt{\left(x-3\right)^2}=\sqrt{\sqrt{3^2}+2.\sqrt{3}.1+1^2}\)
<=> \(\left|x-3\right|=\sqrt{\left(\sqrt{3}+1\right)^2}\)
<=> \(\left|x-3\right|=\sqrt{3}+1\)
=> \(\left[{}\begin{matrix}x-3=\sqrt{3}+1\\x-3=-\left(\sqrt{3}+1\right)\end{matrix}\right.\)=>\(\left[{}\begin{matrix}x=\sqrt{3}+4\\x=-\sqrt{3}+2\end{matrix}\right.\)
Vậy S = \(\left\{\sqrt{3}+4;-\sqrt{3}+2\right\}\)
4) \(\sqrt{3x+2\sqrt{3x}+1}=5\)
<=> \(\sqrt{\sqrt{3x}^2+2.\sqrt{3x}.1+1^2}=5\)
<=> \(\sqrt{\left(\sqrt{3x}+1\right)^2}=5\)
<=> \(\left|\sqrt{3x}+1\right|=5\)
=> \(\left[{}\begin{matrix}\sqrt{3x}+1=5\\\sqrt{3x}+1=-5\end{matrix}\right.\)=> \(\left[{}\begin{matrix}\sqrt{3x}=5-1=4\\\sqrt{3x}=\left(-5\right)-1=-6\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}3x=16\\3x=-6\left(loai\right)\end{matrix}\right.\)=> x = \(\dfrac{16}{3}\) Vậy S = \(\left\{\dfrac{16}{3}\right\}\)
5) \(\sqrt{x^2+2x\sqrt{3}+3}=\sqrt{4-2\sqrt{3}}\)
<=> \(\sqrt{\left(x-\sqrt{3}\right)^2}=\sqrt{\left(\sqrt{3}-1\right)^2}\)
<=> \(\left|x-\sqrt{3}\right|=\sqrt{3}-1\)
<=> \(\left[{}\begin{matrix}x-\sqrt{3}=\sqrt{3}-1\\x-\sqrt{3}=-\left(\sqrt{3}-1\right)\end{matrix}\right.\)=> \(\left[{}\begin{matrix}x=-1\\x=-2\sqrt{3}+1\end{matrix}\right.\)
Vậy S = \(\left\{-1;-2\sqrt{3}+1\right\}\)
6) \(\sqrt{6x+4\sqrt{6x}+4}=7\)
<=> \(\sqrt{\sqrt{6x}^2+2.\sqrt{6x}.2+2^2}=7\)
<=> \(\sqrt{\left(\sqrt{6}+2\right)^2}=7\)
<=> \(\left|\sqrt{6x}+2\right|=7\)
=> \(\left[{}\begin{matrix}\sqrt{6x}+2=7\\\sqrt{6x}+2=-7\end{matrix}\right.\)=>\(\left[{}\begin{matrix}\sqrt{6x}=7-2=5\\\sqrt{6x}=\left(-7\right)-2=-9\left(loai\right)\end{matrix}\right.\)
=> \(\sqrt{6x}=5=>6x=25=>x=\dfrac{25}{6}\)
mầy câu 1;3;;4;5 cách làm nhu nhau(nhân liên hop hoac bình phuong lên)
1.
\(DK:x\in\left[-4;5\right]\)
\(\Leftrightarrow\sqrt{x-5}+\left(\sqrt{x+4}-3\right)=0\)
\(\Leftrightarrow\sqrt{x-5}+\frac{x-5}{\sqrt{x+4}+3}=0\)
\(\Leftrightarrow\sqrt{x-5}\left(1+\frac{\sqrt{x-5}}{\sqrt{x+4}+3}\right)=0\)
Vi \(1+\frac{\sqrt{x-5}}{\sqrt{x+4}+3}>0\)
\(\Rightarrow\sqrt{x-5}=0\)
\(x=5\left(n\right)\)
Vay nghiem cua PT la \(x=5\)
2.
\(DK:x\ge0\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x}-2\right)^2}+\sqrt{\left(\sqrt{x}-3\right)^2}=1\)
\(\Leftrightarrow|\sqrt{x}-2|+|\sqrt{x}-3|=1\)
Ta co:
\(|\sqrt{x}-2|+|\sqrt{x}-3|=|\sqrt{x}-2|+|3-\sqrt{x}|\ge|\sqrt{x}-2+3-\sqrt{x}|=1\)
Dau '=' xay ra khi \(\left(\sqrt{x}-2\right)\left(3-\sqrt{x}\right)\ge0\)
TH1:
\(\hept{\begin{cases}\sqrt{x}-2\ge0\\3-\sqrt{x}\ge0\end{cases}\Leftrightarrow4\le x\le9\left(n\right)}\)
TH2:(loai)
Vay nghiem cua PT la \(x\in\left[4;9\right]\)