Cho M=\(\dfrac{(\sqrt{1+\sqrt{1-x^2})}(\sqrt{\left(1+x^2\right)}-\sqrt{\left(1-x^2\right)})}{2+\sqrt{1-x^2}}\)
Rút gọn M
Cho M=\(\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2\sqrt{x}-2}{x\sqrt{x}-\sqrt{x}+x-1}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{x-1}\right)\)
a) Rút gọn M
b)Tìm GTNN của M
ĐKXĐ: \(x\ge0;x\ne1\)
\(M=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2\sqrt{x}-2}{\left(x-1\right)\left(\sqrt{x}+1\right)}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\left(\dfrac{x-1-2\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(x-1\right)}\right):\left(\dfrac{\sqrt{x}+1-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)^2}.\left(\sqrt{x}+1\right)=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
b.
\(M=\dfrac{\sqrt{x}+1-2}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\ge1-\dfrac{2}{0+1}=-1\)
\(M_{min}=-1\) khi \(x=0\)
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
\(C=\sqrt{m}+\sqrt{m}-1=2\sqrt{m}-1\)
Cho x>1 .Hãy rút gọn đa thức sau:
\(M=1-\left[\left(\dfrac{1}{\sqrt{x}-1}-\sqrt{x-1}\right):\left(\dfrac{1}{\sqrt{x+1}}-\dfrac{1}{\sqrt{x-1}}\right):\dfrac{\sqrt{x}+1.\sqrt{x^2-1}}{\left(x-1\right)\sqrt{x+1}-\left(x+1\right).\sqrt{x-1}}\right]\)
M = \(\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{x-1}\right):\left(\dfrac{2}{x}-\dfrac{2-x}{x\sqrt{x}+x}\right)\)
rút gọn M
tìm X để M = \(\dfrac{-1}{2}\)
Rút gọn \(M=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{\sqrt{x}}{\sqrt{x}+2}\right):\left(\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\right)\)
\(M=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{\sqrt{x}}{\sqrt{x}+2}\right):\left(\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\right)\left(\text{đ}k\text{x}\text{đ}:x\ge0;x\ne4\right)\\ =\left(\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right):\left(\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\right)\\ =\dfrac{x+2\sqrt{x}-\left(x-2\sqrt{x}\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}-1}{x-\sqrt{x}}\\ =\dfrac{x+2\sqrt{x}-x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\\ =\dfrac{4\sqrt{x}}{x-4}\cdot\dfrac{1}{\sqrt{x}}\\ =\dfrac{4\sqrt{x}}{\sqrt{x}\left(x-4\right)}=\dfrac{4}{x-4}\)
\(M=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\right)\div\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{\sqrt{x}}{1-\sqrt{x}}+\dfrac{2}{x-1}\right)\)
a) Rút gọn M
b) Tính M tại \(x=3-2\sqrt{2}\)
c) Tìm x đẻ M = \(\dfrac{8}{9}\)
d) CM \(M\ge0\)
\(M=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{2+5\sqrt{x}}{4-x}\)
\(N=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a+1}}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
Rút gọn :
Với `x >= 0,x ne 4` có:
`M=[(\sqrt{x}+1)(\sqrt{x}+2)+2\sqrt{x}(\sqrt{x}-2)-2-5\sqrt{x}]/[(\sqrt{x}-2)(\sqrt{x}+2)]`
`M=[x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}]/[(\sqrt{x}-2)(\sqrt{x}+2)]`
`M=[3x-6\sqrt{x}]/[(\sqrt{x}-2)(\sqrt{x}+2)]=[3\sqrt{x}]/[\sqrt{x}+2]`
____________
`N=(1/[\sqrt{a}-1]-1/\sqrt{a}):([\sqrt{a}+1]/[\sqrt{a}-2]-[\sqrt{a}+2]/[\sqrt{a}-1])`
- Biểu thức `N` là như vầy?
Với `a > 0,a ne 1,a ne 4` có:
`N=[\sqrt{a}-\sqrt{a}+1]/[\sqrt{a}(\sqrt{a}-1)]:[(\sqrt{a}+1)(\sqrt{a}-1)-(\sqrt{a}+2)(\sqrt{a}-2)]/[(\sqrt{a}-2)(\sqrt{a}-1)]`
`N=1/[\sqrt{a}(\sqrt{a}-1)].[(\sqrt{a}-2)(\sqrt{a}-1)]/[a-1-a+4]`
`N=[\sqrt{a}-2]/[3\sqrt{a}]`
Với \(x\ge0;x\ne4\)
Khi đó:
\(M=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{x-4}+\dfrac{2\sqrt{x}\left(\sqrt{x}-2\right)}{x-4}-\dfrac{2+5\sqrt{x}}{x-4}\\ =\dfrac{x+2\sqrt{x}+\sqrt{x}+2}{x-4}+\dfrac{2x-4\sqrt{x}}{x-4}-\dfrac{2+5\sqrt{x}}{x-4}\\ =\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{x-4}\\ =\dfrac{3x-6\sqrt{x}}{x-4}\\ =\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\\ =\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)
Với \(a>0;a\ne1;a\ne4\)
Khi đó:
\(N=(\dfrac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}-1\right)}):\left(\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}-\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\\ =\left(\dfrac{\sqrt{a}}{a-\sqrt{a}}-\dfrac{\sqrt{a}-1}{a-\sqrt{a}}\right):\left(\dfrac{a-1}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}-\dfrac{a-4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\\ =\dfrac{1}{a-\sqrt{a}}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\\ =\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\\ =\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}-1\right).3}\\ =\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
Cho B=\(\left(\dfrac{4\sqrt{x}}{2+\sqrt{x}}-\dfrac{8x}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\)
a)Rút gọn B
b)Tìm m để với mọi giá trị x>9 ta có \(m\left(\sqrt{x}-3\right)B>x+1\)
a: \(=\dfrac{4x-8\sqrt{x}+8x}{x-4}:\dfrac{\sqrt{x}-1-2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{4\sqrt{x}\left(3\sqrt{x}-2\right)}{x-4}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{-\sqrt{x}+3}=\dfrac{-4x\left(3\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)
b: \(m\left(\sqrt{x}-3\right)\cdot B>x+1\)
=>\(-4xm\left(3\sqrt{x}-2\right)>\left(\sqrt{x}+2\right)\cdot\left(x+1\right)\)
=>\(-12m\cdot x\sqrt{x}+8xm>x\sqrt{x}+2x+\sqrt{x}+2\)
=>\(x\sqrt{x}\left(-12m-1\right)+x\left(8m-2\right)-\sqrt{x}-2>0\)
Để BPT luôn đúng thì m<-0,3
Rút gọn:
1) \(P=\left(\dfrac{\sqrt{x}}{2}-\dfrac{1}{2\sqrt{x}}\right):\left(\dfrac{x-\sqrt{x}}{\sqrt{x}+1}-\dfrac{x-\sqrt{x}}{\sqrt{x}+1}-\dfrac{x+\sqrt{x}}{\sqrt{x}-1}\right)\)
2) \(P=\left(\dfrac{1}{\sqrt{x}}-\dfrac{1}{\sqrt{x}-1}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)
Giúp mk nhé :3
Rút gọn:
\(M=1-\left[\dfrac{2x-1+\sqrt{x}}{1-x}+\dfrac{2x\sqrt{x}+x-\sqrt{x}}{1+x\sqrt{x}}\right]\cdot\left[\dfrac{\left(x-\sqrt{x}\right)\left(1-\sqrt{x}\right)}{2\sqrt{x}-1}\right]\)
Giải::
ĐK: x khác +- 1
\(M=1-\left[\dfrac{\left(\sqrt{x}-\dfrac{1}{2}\right)\left(\sqrt{x}+1\right)}{\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-\dfrac{1}{2}\right)\left(\sqrt{x}+1\right)}{\left(1+\sqrt{x}\right)\left(1-\sqrt{x}+x\right)}\right]\cdot\left[\dfrac{-\sqrt{x}\left(1-\sqrt{x}\right)^2}{2\left(\sqrt{x}-\dfrac{1}{2}\right)}\right]\)
\(=1-\left[\dfrac{\left(\sqrt{x}-\dfrac{1}{2}\right)}{\left(1-\sqrt{x}\right)}\cdot\dfrac{-\sqrt{x}\left(1-\sqrt{x}\right)^2}{2\left(\sqrt{x}-\dfrac{1}{2}\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-\dfrac{1}{2}\right)}{1-\sqrt{x}+x}\cdot\dfrac{-\sqrt{x}\left(1-\sqrt{x}\right)^2}{2\left(\sqrt{x}-\dfrac{1}{2}\right)}\right]\)
\(=1-\left[\dfrac{-\sqrt{x}\left(1-\sqrt{x}\right)}{2}+\dfrac{-x\left(1-\sqrt{x}\right)^2}{2\left(1-\sqrt{x}+x\right)}\right]\)
rồi làm sao nữa ak?? Tớ có quy đồng lên, tính sơ sơ rồi nhưng thấy kq không gọn.
Câu b là : tìm các số nguyên x để M cũng là số nguyên . Nên tớ nghĩ kq sẽ gọn.
NHỜ MẤY CAO NHÂN RA TAY GIÚP VỚI NHAK ^^!