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30 tháng 6 2021
c) Ta có: \(C=\left(\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\right):\dfrac{\sqrt{x}}{x-4}\)
\(=\dfrac{\sqrt{x}-2+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}}=2\)
d)
Sửa đề: \(D=\dfrac{8+x\left(1+\sqrt{x-2\sqrt{x}+1}\right)}{\left(x-4\right)\left(x-2\sqrt{x}+4\right)}+\dfrac{x-3\sqrt{x}}{2\left(x-\sqrt{x}-6\right)}\)
Ta có: \(D=\dfrac{8+x\left(1+\sqrt{x-2\sqrt{x}+1}\right)}{\left(x-4\right)\left(x-2\sqrt{x}+4\right)}+\dfrac{x-3\sqrt{x}}{2\left(x-\sqrt{x}-6\right)}\)
\(=\dfrac{8+x\left(1+\sqrt{x}-1\right)}{\left(x-4\right)\left(x-2\sqrt{x}+4\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x\sqrt{x}+8}{\left(x-4\right)\left(x-2\sqrt{x}+4\right)}+\dfrac{\sqrt{x}}{2\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}}{2\left(\sqrt{x}+2\right)}\)
\(=\dfrac{1}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{2\left(\sqrt{x}+2\right)}\)
\(=\dfrac{2\left(\sqrt{x}+2\right)+\sqrt{x}\left(\sqrt{x}-2\right)}{2\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{2\sqrt{x}+4+x-2\sqrt{x}}{2\left(x-4\right)}\)
\(=\dfrac{x+4}{2x-8}\)
T
27 tháng 4 2020
f) ĐKXĐ: \(x\ge-\frac{3}{2}\)
Khi đó VT > 0 nên \(VT>0\Rightarrow\left[{}\begin{matrix}x\ge2\\x\le-3\left(L\right)\end{matrix}\right.\)
Lũy thừa 6 cả 2 vế lên PT tương đương:
\( \left( x-3 \right) \left( {x}^{11}+9\,{x}^{10}+6\,{x}^{9}-142\,{x}^{ 8}-231\,{x}^{7}+1113\,{x}^{6}+2080\,{x}^{5}-4604\,{x}^{4}-6908\,{x}^{3 }+13222\,{x}^{2}+10983\,x-15327 \right) =0\)
Cái ngoặc to vô nghiệm vì nó tương đương:
\(\left( x-2 \right) ^{11}+31\, \left( x-2 \right) ^{10}+406\, \left( x -2 \right) ^{9}+2906\, \left( x-2 \right) ^{8}+12281\, \left( x-2 \right) ^{7}+31031\, \left( x-2 \right) ^{6}+46656\, \left( x-2 \right) ^{5}+46648\, \left( x-2 \right) ^{4}+46452\, \left( x-2 \right) ^{3}+44590\, \left( x-2 \right) ^{2}+36015\,x-55223 = 0\)(vô nghiệm với mọi \(x\ge2\))
Vậy x = 3.
PS: Nghiệm đẹp thế này chắc có cách AM-Gm độc đáo nhưng mình chưa nghĩ ra
25 tháng 4 2020
@Akai Haruma, @Nguyễn Việt Lâm
giúp em vs ạ! Cần gấp ạ
em cảm ơn nhiều!
25 tháng 8 2021
a)√x−1=2(x≥1)
\(x-1=4
\)
x=5
b)
\(\sqrt{3-x}=4\) (x≤3)
\(\left(\sqrt{3-x}\right)^2=4^2\)
x-3=16
x=19
26 tháng 8 2021
a: Ta có: \(\sqrt{x-1}=2\)
\(\Leftrightarrow x-1=4\)
hay x=5
b: Ta có: \(\sqrt{3-x}=4\)
\(\Leftrightarrow3-x=16\)
hay x=-13
c: Ta có: \(2\cdot\sqrt{3-2x}=\dfrac{1}{2}\)
\(\Leftrightarrow\sqrt{3-2x}=\dfrac{1}{4}\)
\(\Leftrightarrow-2x+3=\dfrac{1}{16}\)
\(\Leftrightarrow-2x=-\dfrac{47}{16}\)
hay \(x=\dfrac{47}{32}\)
d: Ta có: \(4-\sqrt{x-1}=\dfrac{1}{2}\)
\(\Leftrightarrow\sqrt{x-1}=\dfrac{7}{2}\)
\(\Leftrightarrow x-1=\dfrac{49}{4}\)
hay \(x=\dfrac{53}{4}\)
e: Ta có: \(\sqrt{x-1}-3=1\)
\(\Leftrightarrow\sqrt{x-1}=4\)
\(\Leftrightarrow x-1=16\)
hay x=17
f:Ta có: \(\dfrac{1}{2}-2\cdot\sqrt{x+2}=\dfrac{1}{4}\)
\(\Leftrightarrow2\cdot\sqrt{x+2}=\dfrac{1}{4}\)
\(\Leftrightarrow\sqrt{x+2}=\dfrac{1}{8}\)
\(\Leftrightarrow x+2=\dfrac{1}{64}\)
hay \(x=-\dfrac{127}{64}\)
B
19 tháng 8 2018
\(\dfrac{\left(\sqrt{x^2+4}-2\right)\left(\sqrt{x^2+4}+2\right)\left(x+\sqrt{x}+1\right)\sqrt{x-2\sqrt{x}+1}}{x\left(x\sqrt{x}+1\right)}\)
\(=\dfrac{\left[\left(\sqrt{x^2+4}\right)^2-2^2\right]\left(x+\sqrt{x}+1\right)\left(\sqrt{\left(\sqrt{x}-1\right)}\right)^2}{x\left(x\sqrt{x}+1\right)}\)
\(=\dfrac{\left[\left(x^2+4\right)-4\right]\left(x+\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{x\left(x\sqrt{x}+1\right)}\)
\(=\dfrac{\left(x^2+4-4\right)\left(\sqrt{x}^3-1\right)}{x\left(\sqrt{x}^3+1\right)}\)
\(=\dfrac{x^2\left(\sqrt{x}^3-1\right)}{x\left(\sqrt{x}^3+1\right)}\)
\(=\dfrac{x\left(\sqrt{x}^3-1\right)}{\sqrt{x}^3+1}\)
30 tháng 8 2023
1:
\(\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}-2\right)=0\)
=>x-3=0 hoặc \(\sqrt{x+3}=2\)
=>x=3 hoặc x+3=4
=>x=1(loại) hoặc x=3(nhận)
2:
\(\Leftrightarrow\left(\sqrt{4x+1}-\sqrt{3x-4}\right)^2=1\)
=>\(4x-1+3x-4-2\sqrt{\left(4x+1\right)\left(3x-4\right)}=1\)
=>\(\sqrt{4\left(4x+1\right)\left(3x-4\right)}=7x-6\)
=>4(12x^2-16x+3x-4)=(7x-6)^2
=>49x^2-84x+36=48x^2-52x-16
=>-84x+36=-52x-16
=>-32x=-52
=>x=13/8
3: =>\(\sqrt{\left(x-5\right)^2}=5-x\)
=>|x-5|=5-x
=>x-5<=0
=>x<=5
4: \(\Leftrightarrow\left|x-4\right|=x+2\)
=>\(\left\{{}\begin{matrix}x>=-2\\\left(x-4\right)^2=\left(x+2\right)^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=-2\\x^2-8x+16=x^2+4x+4\end{matrix}\right.\)
=>x>=-2 và -8x+16=4x+4
=>x=1
27 tháng 9 2020
a) Ta có: \(B=\left(\sqrt{x}-1\right)\left(\sqrt{x}+4\right)-\left(\sqrt{x}-3\right)^2+\left(2\sqrt{x}+1\right)^2\)
\(=x+4\sqrt{x}-\sqrt{x}-4-\left(x-6\sqrt{x}+9\right)+\left(4x+4\sqrt{x}+1\right)\)
\(=x+3\sqrt{x}-4-x+6\sqrt{x}-9+4x+4\sqrt{x}+1\)
\(=4x+13\sqrt{x}-12\)
b) Ta có: \(C=\frac{x\sqrt{x}+1}{x-\sqrt{x}+1}+\frac{x-4}{x-2}-\frac{x+2\sqrt{x}+1}{\sqrt{x}+1}\)
\(=\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+\frac{x-4}{x-2}-\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}+1}\)
\(=\sqrt{x}+1+\frac{x-4}{x-2}-\left(\sqrt{x}+1\right)\)
\(=\frac{x-4}{x-2}\)
\(x^2-x-4=2\sqrt{x-1}\left(1-x\right)\)
\(ĐKXĐ:x\ge1\)
\(\Leftrightarrow x\left(x-1\right)-4=-2\sqrt{x-1}^3\)
Đặt \(\sqrt{x-1}=a\Rightarrow x=a^2+1\)
\(\Leftrightarrow\left(a^2+1\right)a^2-4=-2a^3\)
\(\Leftrightarrow a^4+2a^3+a^2-4=0\)
\(\Leftrightarrow a^4-a^3+3a^3-3a^2+4a^2-4=0\)(Lm tiếp nha bn.N0 = 1 đó)