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\(\Leftrightarrow\sqrt{4-\left(1-x\right)^2}=\sqrt{3}\)
\(\Leftrightarrow4-\left(1-x\right)^2=3\)
\(\Leftrightarrow4-\left(1-2x+x^2\right)-3=0\)
\(\Leftrightarrow4-1+2x-x^2-3=0\)
\(\Leftrightarrow-x\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}}\)
vay x=0 ; x=2
\(\sqrt{3x^2-5=2}\left(x\ge\sqrt{\frac{5}{3}}\right)\)
\(\Leftrightarrow3x^2-5=4\)
\(\Leftrightarrow3x^2=9\Leftrightarrow x^2=3\Leftrightarrow\orbr{\begin{cases}x=\sqrt{3}\left(tm\right)\\x=-\sqrt{3}\left(kotm\right)\end{cases}}\)
vay \(x=\sqrt{3}\)
\(\sqrt{\left(\sqrt{x}-7\right)\left(\sqrt{x}+7\right)}=2\left(x\ge49\right)\)
\(\Leftrightarrow\sqrt{x-49}=2\Leftrightarrow x^2-98x+2401=4\)
\(\Leftrightarrow x^2-98x+2397=0\Leftrightarrow x^2-47x-51x+2397\)\(\Leftrightarrow x\left(x-47\right)-51\left(x-47\right)\Leftrightarrow\left(x-47\right)\left(x-51\right)\)
\(\Leftrightarrow\orbr{\begin{cases}x-51=0\\x-47=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=51\left(tm\right)\\x=47\left(kotm\right)\end{cases}}}\)
xay x=51
\(\sqrt{\frac{-6}{1+x}}=5\left(x< -1\right)\)
\(\Leftrightarrow\frac{36}{x^2+2x+1}=25\Leftrightarrow25x^2+50x+25=36\)
\(\Leftrightarrow25x^2+50x-11=0\Leftrightarrow25x^2-5x+55x-11\)
\(\Leftrightarrow5x\left(5x-1\right)+11\left(5x-1\right)\Leftrightarrow\left(5x-1\right)\left(5x+11\right)\)\(\Leftrightarrow\orbr{\begin{cases}5x-1=0\\5x+11=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{5}\left(kotm\right)\\x=\frac{-11}{5}\left(tm\right)\end{cases}}}\)
vay \(x=\frac{-11}{5}\)
nhung cau nay binh phuong len la xong
y 3 xem lai de bai
y 4,7 ko biet lam
câu đầu bạn xem lại đề đi nha
các phần còn lại
b)B=\(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}=\sqrt{7-2\sqrt{7}+1}-\sqrt{7+2\sqrt{7}+1}\)=\(\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}\)=\(\sqrt{7}-1-\left(\sqrt{7}+1\right)=-2\)
c)tính từng căn nha
\(\sqrt{13-4\sqrt{3}}=\sqrt{12-2\sqrt{12}+1}=\sqrt{\left(\sqrt{12}-1\right)^2}=\sqrt{12}-1=2\sqrt{3}-1\)
\(\sqrt{22-12\sqrt{2}}=\sqrt{18-4\sqrt{18}+4}=\sqrt{\left(\sqrt{18}-2\right)^2}=\sqrt{18}-2=3\sqrt{2}-3\)
\(\sqrt{\left(2\sqrt{3}-3\sqrt{2}\right)^2}=3\sqrt{2}-2\sqrt{3}\)
thay vào tính C đc C=2
d)có \(\sqrt{9+4\sqrt{2}}=\sqrt{8+2\sqrt{8}+1}=\sqrt{\left(\sqrt{8}+1\right)^2}=\sqrt{8}+1\)\(\Rightarrow6\sqrt{2+\sqrt{9+4\sqrt{2}}}=6\sqrt{2+\sqrt{8}+1}=6\sqrt{2+2\sqrt{2}+1}\)
=\(6\sqrt{\left(\sqrt{2}+1\right)^2}=6\left(\sqrt{2}+1\right)=6\sqrt{2}+6\)\(\Rightarrow D=\sqrt{17-6\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{17-6\sqrt{2}-6}=\sqrt{11-6\sqrt{2}}=\sqrt{9-6\sqrt{2}+2}\)
=\(\sqrt{\left(3-\sqrt{2}\right)^2}=3-\sqrt{2}\)
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
\(\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}+\frac{5\left(2\sqrt{2}+\sqrt{3}\right)}{\left(2\sqrt{2}+\sqrt{3}\right)\left(2\sqrt{2}-\sqrt{3}\right)}-\frac{5\left(\sqrt{8}-\sqrt{3}\right)}{\left(\sqrt{8}-\sqrt{3}\right)\left(\sqrt{8}+\sqrt{3}\right)}\)
\(=\sqrt{3}+1+\sqrt{3}-1+\frac{5\left(2\sqrt{2}+\sqrt{3}\right)}{5}-\frac{5\left(\sqrt{8}-\sqrt{3}\right)}{5}\)
\(=2\sqrt{3}+2\sqrt{2}+\sqrt{3}-\sqrt{8}+\sqrt{3}\)
\(=4\sqrt{3}\)
Giải pt:
1/ \(\Leftrightarrow2x-1=5\)
\(\Leftrightarrow2x=6\Rightarrow x=3\)
2/ \(\Leftrightarrow\sqrt{3}x^2=\sqrt{12}\Leftrightarrow x^2=\sqrt{4}=2\)
\(\Leftrightarrow x=\pm\sqrt{2}\)
3/ \(\Leftrightarrow x-5=9\Rightarrow x=14\)
4/ Đề thiếu
5/ \(\Leftrightarrow\left|x-3\right|=9\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=9\\x-3=-9\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-6\end{matrix}\right.\)
6/ \(\Leftrightarrow2\left|1-x\right|=6\)
\(\Leftrightarrow\left|1-x\right|=3\Leftrightarrow\left[{}\begin{matrix}1-x=3\\1-x=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=4\end{matrix}\right.\)
7/ \(\Leftrightarrow9\left(x-1\right)=21^2\)
\(\Leftrightarrow x-1=49\Rightarrow x=50\)
8/ \(\Leftrightarrow x+1=2^3=8\)
\(\Rightarrow x=7\)
9/ \(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\Leftrightarrow\left|2x+1\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=-\frac{7}{2}\end{matrix}\right.\)
10/ \(\Leftrightarrow\sqrt{2}x=\sqrt{50}\Leftrightarrow x=\sqrt{25}\Rightarrow x=5\)
11/ \(\Leftrightarrow\left|2x-1\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
12/ \(\Leftrightarrow3-2x=\left(-2\right)^3=-8\)
\(\Leftrightarrow2x=11\Rightarrow x=\frac{11}{2}\)
22,
1, Đặt √(3-√5) = A
=> √2A=√(6-2√5)
=> √2A=√(5-2√5+1)
=> √2A=|√5 -1|
=> A=\(\dfrac{\sqrt{5}-1}{\text{√2}}\)
=> A= \(\dfrac{\sqrt{10}-\sqrt{2}}{2}\)
2, Đặt √(7+3√5) = B
=> √2B=√(14+6√5)
=> √2B=√(9+2√45+5)
=> √2B=|3+√5|
=> B= \(\dfrac{3+\sqrt{5}}{\sqrt{2}}\)
=> B= \(\dfrac{3\sqrt{2}+\sqrt{10}}{2}\)
3,
Đặt √(9+√17) - √(9-√17) -\(\sqrt{2}\)=C
=> √2C=√(18+2√17) - √(18-2√17) -\(2\)
=> √2C=√(17+2√17+1) - √(17-2√17+1) -\(2\)
=> √2C=√17+1- √17+1 -\(2\)
=> √2C=0
=> C=0
26,
|3-2x|=2\(\sqrt{5}\)
TH1: 3-2x ≥ 0 ⇔ x≤\(\dfrac{-3}{2}\)
3-2x=2\(\sqrt{5}\)
-2x=2\(\sqrt{5}\) -3
x=\(\dfrac{3-2\sqrt{5}}{2}\) (KTMĐK)
TH2: 3-2x < 0 ⇔ x>\(\dfrac{-3}{2}\)
3-2x=-2\(\sqrt{5}\)
-2x=-2√5 -3
x=\(\dfrac{3+2\sqrt{5}}{2}\) (TMĐK)
Vậy x=\(\dfrac{3+2\sqrt{5}}{2}\)
2, \(\sqrt{x^2}\)=12 ⇔ |x|=12 ⇔ x=12, -12
3, \(\sqrt{x^2-2x+1}\)=7
⇔ |x-1|=7
TH1: x-1≥0 ⇔ x≥1
x-1=7 ⇔ x=8 (TMĐK)
TH2: x-1<0 ⇔ x<1
x-1=-7 ⇔ x=-6 (TMĐK)
Vậy x=8, -6
4, \(\sqrt{\left(x-1\right)^2}\)=x+3
⇔ |x-1|=x+3
TH1: x-1≥0 ⇔ x≥1
x-1=x+3 ⇔ 0x=4 (KTM)
TH2: x-1<0 ⇔ x<1
x-1=-x-3 ⇔ 2x=-2 ⇔x=-1 (TMĐK)
Vậy x=-1