a) \(\sqrt{9x^2}=6\Leftrightarrow\left|x\right|=2\Rightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)

b) \(\sqrt{\left(x-2\right)^2}=5\Leftrightarrow\left|x-2\right|=5\Leftrightarrow\orbr{\begin{cases}x-2=5\\x-2=-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=7\\x=-3\end{cases}}\)

c) \(\sqrt{x^2-6x+9}=3\Leftrightarrow\left|x-3\right|=3\Leftrightarrow\orbr{\begin{cases}x-3=3\\x-3=-3\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=6\\x=0\end{cases}}\)

d) \(\sqrt{x^2+4x+4}-2x=3\Leftrightarrow\left|x+2\right|=2x+3\Leftrightarrow\orbr{\begin{cases}x+2=2x+3\\x+2=-2x-3\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-1\\3x=-5\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\x=-\frac{5}{3}\end{cases}}\)

\(a,\sqrt{9x^2}=6\)

\(\sqrt{\left(3x\right)^2}=6\)

\(\left|3x\right|=6\)

\(< =>x=\pm2\)

\(\sqrt{\left(x-2\right)^2}=5\)

\(\left|x-2\right|=5\)

\(\orbr{\begin{cases}x-2=5\\x-2=-5\end{cases}\orbr{\begin{cases}x=7\\x=-3\end{cases}}}\)

\(c,\sqrt{x^2-6x+9}=3\)

\(\sqrt{\left(x-3\right)^2}=3\)

\(\left|x-3\right|=3\)

\(\orbr{\begin{cases}x-3=3\\x-3=-3\end{cases}\orbr{\begin{cases}x=6\\x=0\end{cases}}}\)

\(d,\sqrt{x^2+4x+4}-2x=3\)

\(\sqrt{\left(x+2\right)^2}=3+2x\)

\(\left|x+2\right|=3+2x\)

dạng 3 của trị tuyệt đối

xét \(3+2x\ge0\)

\(x\ge-\frac{3}{2}\)

\(\orbr{\begin{cases}x+2=3+2x\\x+2=-3-2x\end{cases}\orbr{\begin{cases}x=-1\\-\frac{5}{3}\end{cases}}}\)

\(A=2\sqrt{\left(-3\right)^6}+2\sqrt{\left(-2\right)^4}-4\sqrt{\left(-2\right)^6}\)

\(=2\cdot3^3+2\cdot2^2-4\cdot2^3=30\)

\(B=\sqrt{\left(\sqrt{2}-2\right)^2}+\sqrt{\left(\sqrt{2}-3\right)^2}\)

\(=2-\sqrt{2}+3-\sqrt{2}=5-2\sqrt{2}\)

\(C=\sqrt{\left(3-\sqrt{3}\right)^2}-\sqrt{\left(1+\sqrt{3}\right)^2}\)

\(=3-\sqrt{3}-1-\sqrt{3}=2-2\sqrt{3}\)

\(D=\sqrt{\left(5+\sqrt{6}\right)^2}=5+\sqrt{6}\)

\(E=\sqrt{17^2}-8^2+\sqrt{3^2+4^2}=17-64+5=-42\)

\(A=2\sqrt{\left(-3^3\right)^2}+2\sqrt{\left(-2^2\right)^2}-4\sqrt{\left(-2^3\right)^2}\)

\(A=2\left|-3^3\right|+2\left|-2^2\right|-4\left|-2^3\right|\)

\(2.27+2.4-4.8\)

\(A=30\)

\(B=\sqrt{\left(\sqrt{2}-2\right)^2}+\sqrt{\left(\sqrt{2}-3\right)^2}\)

\(B=\left|\sqrt{2}-2\right|+\left|\sqrt{2}-3\right|\)

\(B=2-\sqrt{2}+3-\sqrt{2}\)

\(B=5-2\sqrt{2}\)

\(C=\sqrt{\left(3-\sqrt{3}\right)^2}-\sqrt{\left(1+\sqrt{3}\right)^2}\)

\(C=\left|3-\sqrt{3}\right|-\left|1+\sqrt{3}\right|\)

\(C=3-\sqrt{3}-1-\sqrt{3}\)

\(C=2-2\sqrt{3}\)

\(D=\sqrt{\left(5+\sqrt{6}\right)^2}\)

\(D=5+\sqrt{6}\)

\(E=\sqrt{17^2}-8^2-\sqrt{3^2+4^2}\)

\(E=17-64-\sqrt{25}=17-64-5=-52\)

Ta có: \(\left(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)

\(=\left[\frac{3\left(6-1\right)}{\sqrt{6}+1}+\frac{2\left(6-4\right)}{\sqrt{6}-2}-\frac{4\left(9-6\right)}{3-\sqrt{6}}\right]\left(\sqrt{6}+11\right)\)

\(=\left[3\left(\sqrt{6}-1\right)+2\left(\sqrt{6}+2\right)-4\left(3+\sqrt{6}\right)\right]\left(\sqrt{6}+11\right)\)

\(=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)

\(=\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)=6-121=-115\)

a) \(2x-4\ge0\Rightarrow x\ge2\)

b) \(-2x+1>0\Rightarrow x< \frac{1}{2}\)

c) \(-3x+5\le0\Rightarrow x\ge\frac{5}{3}\)

d) \(-2x+6\le0\Rightarrow x\ge3\)

e) \(x-3\ge0\Rightarrow x\ge3\)

f) \(-x+2\ge0\Rightarrow x\le2\)

\(a,\sqrt{2x-4}\ge0\)

\(2x-4\ge0\)

\(x\ge2\)để biểu thức ĐXĐ

\(b,\sqrt{\frac{3}{-2x+1}}\ge0\)

\(\frac{3}{-2x+1}\ge0\)

\(< =>-2x+1\ge0;-2x+1\ne0< =>-2x+1>0\)

\(x< \frac{1}{2}\)

\(c,\sqrt{\frac{-3x+5}{-4}}\ge0\)

\(-3x+5\le0\)

\(x\ge\frac{5}{3}\)

mấy câu khác tương tự

a) đk: \(\hept{\begin{cases}x\ge1\\x\ne3\end{cases}}\)

\(P=\frac{x-3}{\sqrt{x-1}-\sqrt{2}}=\frac{\left(x-1\right)-2}{\sqrt{x-1}-\sqrt{2}}\)

\(=\frac{\left(\sqrt{x-1}-\sqrt{2}\right)\left(\sqrt{x-1}+\sqrt{2}\right)}{\sqrt{x-1}-\sqrt{2}}=\sqrt{x-1}+\sqrt{2}\)

b) \(x=4\left(2-\sqrt{3}\right)\Rightarrow x-1=7-4\sqrt{3}=\left(2-\sqrt{3}\right)^2\)

\(\Rightarrow P=\sqrt{x-1}+\sqrt{2}=2-\sqrt{3}+\sqrt{2}\)

Đề 1:

Câu 1: Chọn \(C.\)\(9cm\).

Câu 2: Chọn \(D.\)\(25cm\).

Câu 3: Chọn \(B.\)\(BA^2=BC.BH\).

Câu 4: Chọn \(C.\)\(\sqrt{HB.HC}\).

Đề 2:

Câu 1: Chọn \(B.\)\(8\).

Câu 2: Chọn \(B.\)\(6\).

Câu 3: Chọn \(C.\)\(8\).

Câu 4: Chọn \(A.\)\(\frac{1}{AI^2}=\frac{1}{AD^2}+\frac{1}{4HC^2}\).

- đk : \(x^2-x-2\ge0\Leftrightarrow\left(x-2\right)\left(x+1\right)\ge0\left(1\right)\)

- đặt u = x - 2 => x = u + 2 . Khi đó pt trở thành :

\(\sqrt{u^2+2u}+\sqrt{u^2-3u+4}=2\)

\(\Leftrightarrow2u^2-u+4+2\sqrt{\left(u^2+2u\right)\left(u^2-3u+4\right)}=4\)

\(\Leftrightarrow4\left(u^2+2u\right)\left(u^2-3u+4\right)=\left(u-2u^2\right)^2\)( đk : \(u-2u^2\ge0\)(2))

\(\Leftrightarrow4\left(u^4-u^3-u^2+8u\right)=4u^4-4u^3+u^2\)

\(\Leftrightarrow-5u^2+32u=0\)

\(\Leftrightarrow\orbr{\begin{cases}u=0\\u=\frac{32}{4}\end{cases}}\)( tm (2) )

Với u = 0 ; x = 2 ( tmđk 1 )

Vậy x = 2 là nghiệm duy nhất của pt

đk: \(\orbr{\begin{cases}x\ge2\\x\le-1\end{cases}}\)

Ta có: \(\sqrt{x^2-x-2}+\sqrt{x^2-7x+14}=2\)

\(\Leftrightarrow\frac{x^2-x-2}{\sqrt{x^2-x-2}}+\frac{x^2-7x+14-4}{\sqrt{x^2-7x+14}+2}=0\)

\(\Leftrightarrow\frac{\left(x+1\right)\left(x-2\right)}{\sqrt{x^2-x-2}}+\frac{\left(x-2\right)\left(x-5\right)}{\sqrt{x^2-7x+14}+2}=0\)

\(\Leftrightarrow\left(x-2\right)\left(\frac{x+1}{\sqrt{x^2-x-2}}+\frac{x-5}{\sqrt{x^2-7x+14}+2}\right)=0\)

Xét linh tinh thì cái BT có căn chắc luôn dương hoặc âm

=> x - 2 = 0 => x = 2