rút gọn
√ x^2-4x+4//x-2+√[1-√3]^2
rút gọn
√ x^2-4x+4//x-2+√[1-√3]^2
Em vào phần \(\Sigma\) ở góc bên trái để đăng đề bài cho dễ nhìn nhé
a: \(VT=\sqrt{6^2-\left(\sqrt{35}\right)^2}=\sqrt{36-35}=1=VP\)
b: \(\left(\sqrt{4}-\sqrt{3}\right)^2=4+3-2\cdot\sqrt{4\cdot3}\)
\(=7-2\sqrt{12}\)
\(=\sqrt{49}-\sqrt{48}\)
a, = 3x - 2x = x
b, = 3.( 3 - a ) = 9 - 3a
c, = a.( a - 5 ) = a^2 - 5a
d, = 2a/3 . 3a/2 = a . a = a^2
e, = 1/a-b . a^2 . ( a - b ) = a^2
e chỉ bt rút gọn đến đây th ạ
mong a chj thông cảm ạ
\(a,\sqrt{9x^2}-2x\\ =\left|3x\right|-2x\\ =-3x-2x\left(x< 0\right)\\ =-5x\\ b,\sqrt{9\left(3-a\right)^2}\\ =\left|3\left(3-a\right)\right|\\ =3\left|3-a\right|\\ =3\cdot\left(a-3\right)\\ c,\sqrt{a^2\left(a-5\right)^2}\\ =\left|a\left(a-5\right)\right|\\ =\left|a\right|\cdot\left|a-5\right|\\ =\left(-a\right)\cdot\left(5-a\right)\\ =a^2-5a\)
a) \(A=\dfrac{x}{y}.\sqrt[]{\dfrac{x^2}{y^4}}\) \(\left(x>0;y\ne0\right)\)
\(\Leftrightarrow A=\dfrac{x}{y}.\sqrt[]{\left(\dfrac{x^{ }}{y^2}\right)^2}\)
\(\Leftrightarrow A=\dfrac{x}{y}.\left|\dfrac{x}{y^2}\right|\)
\(\Leftrightarrow A=\dfrac{x}{y}.\dfrac{x}{y^2}=\dfrac{x^2}{y^3}\left(x>0\right)\)
b) \(B=2y^2\sqrt[]{\dfrac{x^4}{4y^2}}\left(y< 0\right)\)
\(\Leftrightarrow B=2y^2\sqrt[]{\left(\dfrac{x^2}{2y^{ }}\right)^2}\)
\(\Leftrightarrow B=2y^2.\left|\dfrac{x^2}{2y^{ }}\right|\)
\(\Leftrightarrow B=2y^2.\dfrac{x^2}{-2y^{ }}\)
\(\Leftrightarrow B=-x^2y^{ }\)
c) \(C=5xy.\sqrt[]{\dfrac{25x^2}{y^6}}\left(x< 0;y>0;y>1\right)\)
\(\Leftrightarrow C=5xy.\sqrt[]{\left(\dfrac{5x^{ }}{y^3}\right)^2}\)
\(\Leftrightarrow C=5xy.\left|\dfrac{5x^{ }}{y^3}\right|\)
\(\Leftrightarrow C=5xy.\dfrac{-5x^{ }}{y^3}\)
\(\Leftrightarrow C=\dfrac{-25x^2}{y^2}\)
d) \(D=\dfrac{x-2}{\sqrt[]{y-1}+1}.\dfrac{\sqrt[]{y+2\sqrt[]{y-1}}}{\sqrt[]{\left(x-2\right)^4}}\left(x\ne2\right)\)
\(\Leftrightarrow D=\dfrac{x-2}{\sqrt[]{y-1}+1}.\dfrac{\sqrt[]{y-1+2\sqrt[]{y-1}+1}}{\left|\left(x-2\right)^2\right|}\)
\(\Leftrightarrow D=\dfrac{x-2}{\sqrt[]{y-1}+1}.\dfrac{\sqrt[]{\left(\sqrt[]{y-1}+1\right)^2}}{\left(x-2\right)^2}\)
\(\Leftrightarrow D=\dfrac{1}{\left(\sqrt[]{y-1}+1\right)}.\dfrac{\left|\sqrt[]{y-1}+1\right|}{x-2}\)
\(\Leftrightarrow D=\dfrac{1}{\left(\sqrt[]{y-1}+1\right)}.\dfrac{\sqrt[]{y-1}+1}{x-2}\)
\(\Leftrightarrow D=\dfrac{1}{x-2}\left(y\ge1;\sqrt[]{y-1}+1>0\right)\)
rút gọn a = 1 phần căn a - 1 - 1 phần căn a đóng ngoặc chia mở ngoặc căn a + 1 phần căn a - 2 - căn a + 2 phần căn a - 1 đóng ngoặc
a. rút gọn a
b. tìm a đẻ A dương
ta có thể làm như sau: Bước 1: Rút gọn phần tử trong ngoặc đầu tiên: √a - 1 - 1 / √a = (√a * √a - √a - 1) / √a = (a - √a - 1) / √a Bước 2: Rút gọn phần tử trong ngoặc thứ hai: √a - 2 - √(a + 2) / √(a - 1) = (√a * √(a - 1) - 2 * √(a - 1) - √(a + 2)) / √(a - 1) = (a - √a - 2√(a - 1) - √(a + 2)) / √(a - 1) Bước 3: Thay các giá trị rút gọn vào biểu thức ban đầu: a = 1 / ((a - √a - 1) / √a) / (√a + 1 / ((a - √a - 2√(a - 1) - √(a + 2)) / √(a - 1))) Bước 4: Rút gọn biểu thức: a = √a * √(a - 1) / (a - √a - 1) * (√(a - 1) / (a - √a - 2√(a - 1) - √(a + 2))) Bước 5: Rút gọn thêm: a = √a * √(a - 1) / (a - √a - 1) * (√(a - 1) / (a - √a - 2√(a - 1) - √(a + 2))) * (√(a - 1) / (a - √a - 2√(a - 1) - √(a + 2))) Bước 6: Rút gọn thêm: a = (√a * √(a - 1))^2 / (a - √a - 1) * (√(a - 1))^2 / (a - √a - 2√(a - 1) - √(a + 2)) Bước 7: Rút gọn cuối cùng: a = (a(a - 1)) / ((a - √a - 1)(a - √a - 2√(a - 1) - √(a + 2)))
\(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}\)
\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}\right)-\sqrt{2}\)
\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{7}+1-\sqrt{7}+1\right)-\sqrt{2}\)
\(=\sqrt{2}-\sqrt{2}=0\)
\(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=1+\sqrt{2}\)
\(1+\dfrac{\sqrt{\left(x-1\right)}}{x-1}\)
=(x-1)/(x-1) + \(\sqrt{x-1}\)/(x-1)
=\(\dfrac{x-1+\sqrt{x-1}}{x-1}\)
=\(\dfrac{\sqrt{x-1}\left(\sqrt{x-1}+1\right)}{\sqrt{\left(x-1\right)^2}}\)
=\(\dfrac{\sqrt{x-1}+1}{\sqrt{x-1}}\)
\(1+\dfrac{\sqrt{\left(x-1\right)}}{x-1}\)
\(=1+\dfrac{1}{\sqrt{x-1}}=\dfrac{\sqrt{x-1}+1}{\sqrt{x-1}}\)
rút gọn hoạc tính giá trị các biểu thức sau
1)1+\(\sqrt{\dfrac{\left(x-1\right)^2}{x-1}}\)
2)\(\sqrt{\left(x-2\right)^2}+\dfrac{x-2}{\sqrt{\left(x-2\right)^2}}\)
3)\(\sqrt{m}-\sqrt{m-2\sqrt{m}+1}\)
1: \(1+\sqrt{\dfrac{\left(x-1\right)^2}{x-1}}=1+\sqrt{x-1}\)
2: \(A=\sqrt{\left(x-2\right)^2}+\dfrac{x-2}{\sqrt{\left(x-2\right)^2}}\)
=\(\left|x-2\right|+\dfrac{x-2}{\left|x-2\right|}\)
TH1: x>2
A=x-2+(x-2)/(x-2)=x-2+1=x-1
TH2: x<2
A=2-x+(x-2)/(2-x)=2-x-1=1-x
3: \(C=\sqrt{m}-\sqrt{m-2\sqrt{m}+1}\)
\(=\sqrt{m}-\sqrt{\left(\sqrt{m}-1\right)^2}\)
\(=\sqrt{m}-\left|\sqrt{m}-1\right|\)
TH1: m>=1
\(C=\sqrt{m}-\sqrt{m}+1=1\)
TH2: 0<=m<1
\(C=\sqrt{m}+\sqrt{m}-1=2\sqrt{m}-1\)
tìm x để căn thức sau được xác định
1)\(\sqrt{\dfrac{-2}{2x-2}}\)
2)\(\sqrt{\dfrac{2}{3x-1}}\)
3)\(\sqrt{\dfrac{2x-2}{-2}}\)
4)\(\sqrt{\dfrac{3x-2}{5}}\)
5)\(\sqrt{\dfrac{x-2}{x+3}}\)
1: ĐKXĐ: -2/2x-2>=0
=>2x-2<0
=>x<1
2: ĐKXĐ: 2/3x-1>=0
=>3x-1>0
=>x>1/3
3: ĐKXĐ: 2x-2/(-2)>=0
=>2x-2<=0
=>x<=1
4: ĐKXĐ: (3x-2)/5>=0
=>3x-2>=0
=>x>=2/3
5: ĐKXĐ: (x-2)/(x+3)>=0
=>x>=2 hoặc x<-3