e: \(f\left(-x\right)=\dfrac{\left(-x\right)^4+3\cdot\left(-x\right)^2-1}{\left(-x\right)^2-4}=\dfrac{x^4+3x^2-1}{x^2-4}=f\left(x\right)\)

Vậy: f(x) là hàm số chẵn

\(c,f\left(-x\right)=\sqrt{-2x+9}=-f\left(x\right)\)

Vậy hàm số lẻ

\(d,f\left(-x\right)=\left(-x-1\right)^{2010}+\left(1-x\right)^{2010}\\ =\left[-\left(x+1\right)\right]^{2010}+\left(x-1\right)^{2010}\\ =\left(x+1\right)^{2010}+\left(x-1\right)^{2010}=f\left(x\right)\)

Vậy hàm số chẵn

\(g,f\left(-x\right)=\sqrt[3]{-5x-3}+\sqrt[3]{-5x+3}\\ =-\sqrt[3]{5x+3}-\sqrt[3]{5x-3}=-f\left(x\right)\)

Vậy hàm số lẻ

\(h,f\left(-x\right)=\sqrt{3-x}-\sqrt{3+x}=-f\left(x\right)\)

Vậy hàm số lẻ

a)đk:`2x-4>=0`

`<=>2x>=4`

`<=>x>=2.`

b)đk:`3/(-2x+1)>=0`

Mà `3>0`

`=>-2x+1>=0`

`<=>1>=2x`

`<=>x<=1/2`

c)`đk:(-3x+5)/(-4)>=0`

`<=>(3x-5)/4>=0`

`<=>3x-5>=0`

`<=>3x>=5`

`<=>x>=5/3`

d)`đk:-5(-2x+6)>=0`

`<=>-2x+6<=0`

`<=>2x-6>=0`

`<=>2x>=6`

`<=>x>=3`

e)`đk:(x^2+2)(x-3)>=0`

Mà `x^2+2>=2>0`

`<=>x-3>=0`

`<=>x>=3`

f)`đk:(x^2+5)/(-x+2)>=0`

Mà `x^2+5>=5>0`

`<=>-x+2>0`

`<=>-x>=-2`

`<=>x<=2`

a, ĐKXĐ : \(2x-4\ge0\)

\(\Leftrightarrow x\ge\dfrac{4}{2}=2\)

Vậy ..

b, ĐKXĐ : \(\left\{{}\begin{matrix}\dfrac{3}{-2x+1}\ge0\\-2x+1\ne0\end{matrix}\right.\)

\(\Leftrightarrow-2x+1>0\)

\(\Leftrightarrow x< \dfrac{1}{2}\)

Vậy ..

c, ĐKXĐ : \(\dfrac{-3x+5}{-4}\ge0\)

\(\Leftrightarrow-3x+5\le0\)

\(\Leftrightarrow x\ge\dfrac{5}{3}\)

Vậy ...

d, ĐKXĐ : \(-5\left(-2x+6\right)\ge0\)

\(\Leftrightarrow-2x+6\le0\)

\(\Leftrightarrow x\ge-\dfrac{6}{-2}=3\)

Vậy ...

e, ĐKXĐ : \(\left(x^2+2\right)\left(x-3\right)\ge0\)

\(\Leftrightarrow x-3\ge0\)

\(\Leftrightarrow x\ge3\)

Vậy ...

f, ĐKXĐ : \(\left\{{}\begin{matrix}\dfrac{x^2+5}{-x+2}\ge0\\-x+2\ne0\end{matrix}\right.\)

\(\Leftrightarrow-x+2>0\)

\(\Leftrightarrow x< 2\)

Vậy ...

a,\(\sqrt{x+6-4\sqrt{x+2}}+\sqrt{x+11-6\sqrt{x+2}}=1\) (*)(đk \(x\ge-2\))

<=> \(\sqrt{\left(x+2\right)-4\sqrt{x+2}+4}+\sqrt{\left(x+2\right)-6\sqrt{x+2}+9}\)=1

<=> \(\sqrt{\left(\sqrt{x+2}-2\right)^2}+\sqrt{\left(\sqrt{x+2}-3\right)^2}=1\)

<=> \(\left|\sqrt{x+2}-2\right|+\left|\sqrt{x+2}-3\right|\)=1 (1)

TH1: \(0\le\sqrt{x+2}< 2\)

Từ (1) =>\(2-\sqrt{x+2}+3-\sqrt{x+2}=1\)

<=> \(5-2\sqrt{x+2}=1\) <=> \(2\sqrt{x+1}=4\) <=> \(\sqrt{x+1}=2\)

<=> \(x+1=4\) <=> x=3(không t/m \(\sqrt{x+2}\le2\))

TH2 : \(2\le\sqrt{x+2}\le3\)

Từ (1) =>\(\sqrt{x+2}-2+3-\sqrt{x+2}=1\)

<=> \(1=1\) (luôn đúng)

Từ TH2 <=> 4\(\le x+2\le9\) <=> \(2\le x\le7\)

TH3 \(\sqrt{x+2}>3\)

Từ (1) => \(\sqrt{x+2}-2+\sqrt{x+2}-3=1\)

<=> \(2\sqrt{x+2}=6\) <=> \(\sqrt{x+2}=3\) <=> \(x+2=9\) <=> x=7 (không t/m \(\sqrt{x+2}>3\))

Vậy pt (*) có tập nghiệm S=\(\left\{2\le x\le7\right\}\)

b, \(x^2-10x+27=\sqrt{6-x}+\sqrt{x-4}\) (*) (đk :\(4\le x\le6\))

Vs a,b \(\ge0\) ta có \(\sqrt{a}+\sqrt{b}\le\sqrt{2\left(a^2+b^2\right)}\)(tự CM nha)

Dấu "=" xảy ra <=> a=b

Áp dụng bđt trên ta có: \(\sqrt{6-x}+\sqrt{x-4}\le\sqrt{2\left(6-x+x-4\right)}=\sqrt{2.2}=2\)

<=> \(\sqrt{6-x}+\sqrt{x-4}\le2\)(1)

Lại có: \(x^2-10x+27=x^2-10x+25+2=\left(x-5\right)^2+2\ge2\)

<=> \(x^2-10x+27\ge2\) (2)

Từ (1),(2) => Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}6-x=x-4\\x-5=0\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}6+4=2x\\x=5\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=5\\x=5\end{matrix}\right.\left(tm\right)\)

Vậy pt (*) có tập nghiệm S=\(\left\{5\right\}\)

c, \(x^2-2x-x\sqrt{x}-2\sqrt{x}+4=0\)(*) (đk: x\(\ge0\))

<=> \(x\left(x-2\right)-\sqrt{x}\left(x-2\right)-4\left(\sqrt{x}-1\right)=0\)

<=> \(\left(x-\sqrt{x}\right)\left(x-2\right)-4\left(\sqrt{x}-1\right)=0\)

<=> \(\sqrt{x}\left(\sqrt{x}-1\right)\left(x-2\right)-4\left(\sqrt{x}-1\right)=0\)

<=> \(\left(\sqrt{x}-1\right)\left[\sqrt{x}\left(x-2\right)-4\right]=0\)

<=> \(\left[{}\begin{matrix}\sqrt{x}-1=0\\\sqrt{x}\left(x-2\right)-4=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}\sqrt{x}=1\\\sqrt{x}\left(x-2\right)=4\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=1\\x\left(x-2\right)^2=16\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=1\\x\left(x^2-4x+4\right)-16=0\end{matrix}\right.\) <=>\(\left[{}\begin{matrix}x=1\\x^3-4x^2+4x-16=0\end{matrix}\right.\)<=> \(\left[{}\begin{matrix}x=1\\x^2\left(x-4\right)+4\left(x-4\right)=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=1\\\left(x^2+4\right)\left(x-4\right)=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=1\\x-4=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=1\\x=4\end{matrix}\right.\left(tm\right)\)

Vậy pt (*) có tập nghiệm S=\(\left\{1;4\right\}\)

Hung nguyen, Trần Thanh Phương, Sky SơnTùng, @tth_new, @Nguyễn Việt Lâm, @Akai Haruma, @No choice teen

help me, pleaseee

Cần gấp lắm ạ!

a/ ĐKXĐ: ...

\(\Leftrightarrow3\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)-7\)

Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=a>0\Rightarrow a^2=x+\frac{1}{4x}+1\)

\(\Rightarrow x+\frac{1}{4x}=a^2-1\)

Pt trở thành:

\(3a=2\left(a^2-1\right)-7\)

\(\Leftrightarrow2a^2-3a-9=9\Rightarrow\left[{}\begin{matrix}a=3\\a=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)

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

\(\Leftrightarrow2x-6\sqrt{x}+1=0\)

\(\Rightarrow\sqrt{x}=\frac{3+\sqrt{7}}{2}\Rightarrow x=\frac{8+3\sqrt{7}}{2}\)

b/ ĐKXĐ:

\(\Leftrightarrow5\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)+4\)

Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=a>0\Rightarrow x+\frac{1}{4x}=a^2-1\)

\(\Rightarrow5a=2\left(a^2-1\right)+4\Leftrightarrow2a^2-5a+2=0\)

\(\Rightarrow\left[{}\begin{matrix}a=2\\a=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{x}+\frac{1}{2\sqrt{x}}=2\\\sqrt{x}+\frac{1}{2\sqrt{x}}=\frac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x-4\sqrt{x}+1=0\\2x-\sqrt{x}+1=0\left(vn\right)\end{matrix}\right.\)

c/ ĐKXĐ: ...

\(\Leftrightarrow\sqrt{2x^2+8x+5}-4\sqrt{x}+\sqrt{2x^2-4x+5}-2\sqrt{x}=0\)

\(\Leftrightarrow\frac{2x^2-8x+5}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\frac{2x^2-8x+5}{\sqrt{2x^2-4x+5}+2\sqrt{x}}=0\)

\(\Leftrightarrow\left(2x^2-8x+5\right)\left(\frac{1}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\frac{1}{\sqrt{2x^2-4x+5}+2\sqrt{x}}\right)=0\)

\(\Leftrightarrow2x^2-8x+5=0\)

d/ ĐKXĐ: ...

\(\Leftrightarrow x+1-\frac{15}{6}\sqrt{x}+\sqrt{x^2-4x+1}-\frac{1}{2}\sqrt{x}=0\)

\(\Leftrightarrow\frac{x^2-\frac{17}{4}x+1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{x^2-\frac{17}{4}x+1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}=0\)

\(\Leftrightarrow\left(x^2-\frac{17}{4}x+1\right)\left(\frac{1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}\right)=0\)

\(\Leftrightarrow x^2-\frac{17}{4}x+1=0\)

\(\Leftrightarrow4x^2-17x+4=0\)

e/ ĐKXĐ: ...

\(\Leftrightarrow x^2-1+2x\sqrt{\frac{x^2-1}{x}}=3x\)

Nhận thấy \(x=0\) không phải nghiệm, pt tương đương:

\(\frac{x^2-1}{x}+2\sqrt{\frac{x^2-1}{x}}=3\)

Đặt \(\sqrt{\frac{x^2-1}{x}}=a\ge0\)

\(a^2+2a=3\Leftrightarrow a^2+2a-3=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-3\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow\sqrt{\frac{x^2-1}{x}}=1\Leftrightarrow x^2-1=x\Leftrightarrow x^2-x-1=0\)

f/ ĐKXĐ: ...

\(\Leftrightarrow x^2-6+x\sqrt{\frac{x^2-6}{x}}-6x=0\)

Nhận thấy \(x=0\) ko phải nghiệm, pt tương đương:

\(\frac{x^2-6}{x}+\sqrt{\frac{x^2-6}{x}}-6=0\)

Đặt \(\sqrt{\frac{x^2-6}{x}}=a\ge0\)

\(a^2+a-6=0\Rightarrow\left[{}\begin{matrix}a=2\\a=-3\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{\frac{x^2-6}{x}}=2\Leftrightarrow x^2-4x-6=0\)

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

@Akai Haruma, @Nguyễn Việt Lâm

giúp em vs ạ! Cần gấp ạ

em cảm ơn nhiều!