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a. ĐKXĐ: \(4-5x\ge0\) \(\Leftrightarrow-5x\ge-4\Leftrightarrow5x\le4\Leftrightarrow x\le\dfrac{4}{5}\)
\(\sqrt{4-5x}=12\)
\(\Leftrightarrow4-5x=2\sqrt{3}\)
\(\Leftrightarrow-5x=-4-2\sqrt{3}\)
\(\Leftrightarrow x=\dfrac{-4-2\sqrt{3}}{-5}\)
\(\Leftrightarrow x=\dfrac{4+2\sqrt{3}}{5}\left(KTMĐKXĐ\right)\)
Vậy x không tồn tại
b. \(10-2\sqrt{2x+1}=4\) (1)
\(ĐKXĐ:2x+1\ge0\Leftrightarrow2x\ge-1\Leftrightarrow x\ge-\dfrac{1}{2}\)
(1) => \(-2\sqrt{2x+1}=-6\)
\(\Leftrightarrow\sqrt{2x+1}=3\)
\(\Leftrightarrow2x+1=\sqrt{3}\)
\(\Leftrightarrow2x=\sqrt{3}-1\)
\(\Leftrightarrow x=\dfrac{\sqrt{3}-1}{2}\left(TMĐKXĐ\right)\)
c. \(5-\sqrt{x-1}=7\) (1)
ĐKXĐ: \(x-1\ge0\Leftrightarrow x\ge1\)
(1) <=> \(-\sqrt{x-1}=2\) (vô lí)
Vậy không tồn tại x
bài kia làm sai rùi:
a. \(\sqrt{4-5x}=12\) (1)
ĐKXĐ: \(4-5x\ge0\Leftrightarrow x\le\dfrac{4}{5}\)
\(\Leftrightarrow4-5x=144\)
\(\Leftrightarrow5x=-140\)
\(\Leftrightarrow x=-28\left(TMĐKXĐ\right)\)
Vậy phương trình có nghiệm là \(S=\left\{-28\right\}\)
b. \(10-2\sqrt{2x+1}=4\) (1)
ĐKXĐ: \(2x+1\ge0\Leftrightarrow x\ge-\dfrac{1}{2}\)
\(\left(1\right)\Leftrightarrow2\sqrt{2x+1}=6\)
\(\Leftrightarrow\sqrt{2x+1}=3\)
\(\Leftrightarrow2x+1=9\)
\(\Leftrightarrow2x=8\)
\(\Leftrightarrow x=4\left(TMĐKXĐ\right)\)
Vậy phương trình có nghiệm là: \(S=\left\{4\right\}\)
c. Ở dưới làm đúng rồi
d. \(\sqrt{10+\sqrt{3x}}=2+\sqrt{6}\) (1)
ĐKXĐ: \(3x\ge0\Leftrightarrow x\ge0\)
(1) \(\Leftrightarrow10+\sqrt{3x}=\left(2+\sqrt{6}\right)^2\)
\(\Leftrightarrow10+\sqrt{3x}=10+4\sqrt{6}\)
\(\Leftrightarrow\sqrt{3x}=-10+10+4\sqrt{6}\)
\(\Leftrightarrow\sqrt{3x}=4\sqrt{6}\)
\(\Leftrightarrow3x=96\)
\(\Leftrightarrow x=32\left(TMĐKXĐ\right)\)
Vậy phương trình có nghiệm là: \(S=\left\{32\right\}\)
e. \(\sqrt{x+1}+10=2\sqrt{x+1}-2\) (1)
ĐKXĐ: \(x+1\ge0\Leftrightarrow x\ge-1\)
\(\left(1\right)\Leftrightarrow\sqrt{x+1}-2\sqrt{x+1}=-10-2\)
\(\Leftrightarrow-\sqrt{x+1}=-12\)
\(\Leftrightarrow\sqrt{x+1}=12\)
\(\Leftrightarrow x+1=144\)
\(\Leftrightarrow x=143\left(TMĐKXĐ\right)\)
Vậy phương trình có nghiệm là \(S=\left\{143\right\}\)
f. \(\sqrt{16x+32}-5\sqrt{x+2}=-2\) (1)
ĐKXĐ: \(\left[{}\begin{matrix}\sqrt{16x+32\ge0}\\\sqrt{x+2\ge0}\end{matrix}\right.\left[{}\begin{matrix}x\ge-2\\x\ge-2\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\sqrt{16\left(x+2\right)}-5\sqrt{x+2}=-2\)
\(\Leftrightarrow4\sqrt{x+2}-5\sqrt{x+2}=-2\)
\(\Leftrightarrow-\sqrt{x+2}=-2\)
\(\Leftrightarrow\sqrt{x+2}=2\)
\(\Leftrightarrow x+2=4\)
\(\Leftrightarrow x=2\left(TMĐKXĐ\right)\)
Vậy phương trình có nghiệm là \(S=\left\{2\right\}\)
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 \(S=\left\{8\right\}\)
6.
ĐKXĐ: \(x\ge2\)
\(\sqrt{\left(x-1\right)\left(x-2\right)}+\sqrt{x+3}=\sqrt{x-2}+\sqrt{\left(x-1\right)\left(x+3\right)}\)
\(\Leftrightarrow\sqrt{\left(x-1\right)\left(x-2\right)}-\sqrt{x-2}+\sqrt{x+3}-\sqrt{\left(x-1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\sqrt{x-2}\left(\sqrt{x-1}-1\right)-\sqrt{x+3}\left(\sqrt{x-1}-1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-2}-\sqrt{x+3}\right)\left(\sqrt{x-1}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-2}=\sqrt{x+3}\\\sqrt{x-1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=x+3\left(vn\right)\\x=2\end{matrix}\right.\)
4.
ĐKXĐ: \(x\ge4\)
Đặt \(\sqrt{x-4}=t\ge0\Rightarrow x=t^2+4\)
\(\Rightarrow3\left(t^2+4\right)+7t=14t-20\)
\(\Leftrightarrow3t^2-7t+34=0\)
Phương trình vô nghiệm
5.
ĐKXĐ: ...
- Với \(x=0\) ko phải nghiệm
- Với \(x\ne0\Rightarrow\sqrt{x+1}-1\ne0\) , nhân 2 vế của pt cho \(\sqrt{x+1}-1\) và rút gọn ta được:
\(\sqrt{x+1}+2x-5=\sqrt{x+1}-1\)
\(\Leftrightarrow2x=4\Rightarrow x=2\)
Bài 3: \(3\left(\sqrt{2x^2+1}-1\right)=x\left(1+3x+8\sqrt{2x^2+1}\right)\)
\(\Leftrightarrow\left(3-8x\right)\sqrt{2x^2+1}=3x^2+x+3\)
\(\Rightarrow\left(3-8x\right)^2\left(2x^2+1\right)=\left(3x^2+x+3\right)^2\)
\(\Leftrightarrow119x^4-102x^3+63x^2-54x=0\)
\(\Leftrightarrow x\left(7x-6\right)\left(17x^2+9\right)=0\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{6}{7}\end{cases}}\)
Thử lại, ta nhận được \(x=0\)là nghiệm duy nhất của phương trình
a, dk \(1-16x^2\ge0\Leftrightarrow\left(1-4x\right)\left(1+4x\right)\ge0\)
\(\Leftrightarrow-\frac{1}{4}\le x\le\frac{1}{4}\)
b tuong tu
c, \(\sqrt{\left(x-3\right)\left(5-x\right)}\ge0\Leftrightarrow\left(x-3\right)\left(5-x\right)\ge0\Leftrightarrow3\le x\le5\)
d.\(\sqrt{x^2-x+1}>0\)
ma \(x^2-x+1=x^2-2.\frac{1}{2}x+\frac{1}{4}+\frac{3}{4}=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\)
suy ra thoa man vs moi x
a) \(\text{Đ}K\text{X}\text{Đ}:\frac{3}{2}\le x\le\frac{5}{2}\)
Áp dụng BĐT Bunhiacopxki ta có:
\(VT=\sqrt{2x-3}+\sqrt{5-2x}\le\sqrt{2\left(2x-3+5-2x\right)}=2\)
Dấu '=' xảy ra khi \(\sqrt{2x-3}=\sqrt{5-2x}\Leftrightarrow x=2\)
Lại có: \(VP=3x^2-12x+14=3\left(x-2\right)^2+2\ge2\)
Dấu '=' xảy ra khi x=2
Do đó VT=VP khi x=2
b) ĐK: \(x\ge0\). Ta thấy x=0 k pk là nghiệm của pt, chia 2 vế cho x ta có:
\(x^2-2x-x\sqrt{x}-2\sqrt{x}+4=0\Leftrightarrow x-2-\sqrt{x}-\frac{2}{\sqrt{x}}+\frac{4}{x}=0\)
\(\Leftrightarrow\left(x+\frac{4}{x}\right)-\left(\sqrt{x}+\frac{2}{\sqrt{x}}\right)-2=0\)
Đặt \(\sqrt{x}+\frac{2}{\sqrt{x}}=t>0\Leftrightarrow t^2=x+4+\frac{4}{x}\Leftrightarrow x+\frac{4}{x}=t^2-4\), thay vào ta có:
\(\left(t^2-4\right)-t-2=0\Leftrightarrow t^2-t-6=0\Leftrightarrow\left(t-3\right)\left(t+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}t=3\\t=-2\end{cases}}\)
Đối chiếu ĐK của t
\(\Rightarrow t=3\Leftrightarrow\sqrt{x}+\frac{2}{\sqrt{x}}=3\Leftrightarrow x-3\sqrt{x}+2=0\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=1\end{cases}}\)
a) \(\sqrt{x-2}+\sqrt{16x-32}=10\)
\(\Rightarrow\sqrt{x-2}+4\sqrt{x-2}=10\)
\(\Rightarrow5\sqrt{x-2}=10\)
\(\Rightarrow\sqrt{x-2}=2\)
\(\Rightarrow x-2=4\)
\(\Rightarrow x=6\)
b) \(\sqrt{x+\sqrt{2x-1}}=5\sqrt{2}\)
ĐK \(x\ge\dfrac{1}{2}\)
\(\sqrt{x+\sqrt{2x-1}}=5\sqrt{2}\)
\(\left(\sqrt{x+\sqrt{2x-1}}\right)^2=\left(5\sqrt{2}\right)^2\)
\(\left|x+\sqrt{2x-1}\right|=50\)
\(\sqrt{2x-1}=50-x\)
\(\left(\sqrt{2x-1}\right)^2=\left(50-x\right)^2\)
\(\left|2x-1\right|=x^2-100x+2500\)
\(2x-1=x^2-100x+2500\)
\(x=41\)