Tìm x :
a) (2x+1)2=\(\dfrac{1}{9}\)
b)\(\sqrt{x+\dfrac{1}{2}}=\dfrac{1}{4}\)
c) \(\dfrac{x+1}{2x-1}=\dfrac{4}{5}\)
Tìm `ĐKXĐ`:
\(\sqrt{\dfrac{-5}{6+x}}\)
\(\sqrt{\dfrac{-2}{6-x}}\)
\(\sqrt{\dfrac{-x+3}{-6}}\)
\(\sqrt{\dfrac{7x-1}{-9}}\)
\(\sqrt{\dfrac{x+2}{x^2+2x+1}}\)
\(\sqrt{\dfrac{x-2}{x^2-2x+4}}\)
\(a,\dfrac{-5}{x+6}\ge0\\ mà\left(-5< 0\right)\\ \Rightarrow x+6< 0\\ \Rightarrow x< -6\\ b,\dfrac{2}{6-x}\ge0\\ mà\left(2>0\right)\\ \Rightarrow6-x>0\\ \Rightarrow x< 6\\ c,\dfrac{-x+3}{-6}\ge0\\ mà-6< 0\\ \Rightarrow-x+3< 0\\ \Rightarrow x>3\\\)
\(d,\dfrac{7x-1}{-9}\ge0\\mà-9< 0\\ \Rightarrow 7x-1\le0\\ \Rightarrow x\le\dfrac{1}{7}\\ e,\dfrac{x+2}{x^2+2x+1}\ge0\\ mà\left(x^2+2x+1\right)>0\forall x\\ \Rightarrow x+2\ge0\\ \Rightarrow x\ge-2\\ f,\dfrac{x-2}{x^2-2x+4}\ge0\\ mà\left(x^2-2x+4\right)>0\forall x\\ \Rightarrow x-2\ge0\\ \Rightarrow x\ge2\)
Chứng minh : \(x^2-2x+4>0\\ x^2-2x+1+3=\left(x-1\right)^2+3\ge3>0\)
a: ĐKXĐ: \(\dfrac{-5}{x+6}>=0\)
=>x+6<0
=>x<-6
b: ĐKXĐ: (-2)/(6-x)>=0
=>6-x<0
=>x>6
c: ĐKXĐ: (-x+3)/(-6)>=0
=>-x+3<=0
=>-x<=-3
=>x>=3
d: ĐKXĐ: (7x-1)/-9>=0
=>7x-1<=0
=>x<=1/7
e: ĐKXĐ: (x+2)/(x^2+2x+1)>=0
=>x+2>=0
=>x>=-1
f: ĐKXĐ: (x-2)/(x^2-2x+4)>=0
=>x-2>=0
=>x>=2
Giải PT:
a) \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)
b) \(\sqrt{18x-9}-0,5\sqrt{2x-1}+\dfrac{1}{2}\sqrt{25\left(2x-1\right)}+\sqrt{49\left(2x-1\right)}=24\)
c) \(\sqrt{36x-72}-15\sqrt{\dfrac{x-2}{25}}=4\left(5+\sqrt{x-2}\right)\)
d) \(\sqrt{\dfrac{1}{3x+2}}-\dfrac{1}{2}\sqrt{\dfrac{9}{3x+2}}+\sqrt{\dfrac{16}{3x+2}}-5\sqrt{\dfrac{1}{12x+8}}=1\)
e) \(\dfrac{1}{2}\sqrt{\dfrac{49x}{x+2}}-3\sqrt{\dfrac{x}{4x+8}}-\sqrt{\dfrac{x}{x+2}}-\sqrt{5}=0\)
a. ĐKXĐ: $x\geq 1$
PT $\Leftrightarrow \frac{1}{2}\sqrt{x-1}-\frac{3}{2}.\sqrt{9}.\sqrt{x-1}+24.\sqrt{\frac{1}{64}}.\sqrt{x-1}=-17$
$\Leftrightarrow \frac{1}{2}\sqrt{x-1}-\frac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17$
$\Leftrightarrow -\sqrt{x-1}=-17$
$\Leftrightarrow \sqrt{x-1}=17$
$\Leftrightarrow x-1=289$
$\Leftrightarrow x=290$
b. ĐKXĐ: $x\geq \frac{1}{2}$
PT $\Leftrightarrow \sqrt{9}.\sqrt{2x-1}-0,5\sqrt{2x-1}+\frac{1}{2}.\sqrt{25}.\sqrt{2x-1}+\sqrt{49}.\sqrt{2x-1}=24$
$\Leftrightarrow 3\sqrt{2x-1}-0,5\sqrt{2x-1}+2,5\sqrt{2x-1}+7\sqrt{2x-1}=24$
$\Leftrightarrow 12\sqrt{2x-1}=24$
$\Leftrihgtarrow \sqrt{2x-1}=2$
$\Leftrightarrow x=2,5$ (tm)
c. ĐKXĐ: $x\geq 2$
PT $\Leftrightarrow \sqrt{36}.\sqrt{x-2}-15\sqrt{\frac{1}{25}}\sqrt{x-2}=4(5+\sqrt{x-2})$
$\Leftrightarrow 6\sqrt{x-2}-3\sqrt{x-2}=20+4\sqrt{x-2}$
$\Leftrightarrow \sqrt{x-2}=-20< 0$ (vô lý)
Vậy pt vô nghiệm
d. ĐKXĐ: $x>\frac{-2}{3}$
PT $\Leftrightarrow \sqrt{\frac{1}{3x+2}}-\frac{1}{2}\sqrt{9}.\sqrt{\frac{1}{3x+2}}+\sqrt{16}.\sqrt{\frac{1}{3x+2}}-5\sqrt{\frac{1}{4}}\sqrt{\frac{1}{3x+2}}=1$
$\Leftrightarrow \sqrt{\frac{1}{3x+2}}-\frac{3}{2}\sqrt{\frac{1}{3x+2}}+4\sqrt{\frac{1}{3x+2}}-\frac{5}{2}\sqrt{\frac{1}{3x+2}}=1$
$\Leftrightarrow \sqrt{\frac{1}{3x+2}}=1$
$\Leftrightarrow \frac{1}{3x+2}=1$
$\Leftrightarrow 3x+2=1$
$\Leftrightarrow x=-\frac{1}{3}$
a.\(\dfrac{4x-5}{x-1}=2+\dfrac{x}{x-1}\) b.\(\dfrac{7}{x+2}=\dfrac{3}{x-5}\) c.\(\dfrac{2x+5}{2x}-\dfrac{x}{x+5}=0\)
d.\(\dfrac{12x+1}{11x-4}+\dfrac{10x-4}{9}=\dfrac{20x+17}{18}\) e.\(\dfrac{11}{x}=\dfrac{9}{x+1}+\dfrac{2}{x-4}\) f.\(\dfrac{14}{3x-12}-\dfrac{2+x}{x-4}=\dfrac{3}{8-2x}-\dfrac{5}{6}\)
m.\(\dfrac{12}{1-9x^2}=\dfrac{1+3x}{1+3x}-\dfrac{1+3x}{1-3x}\) n.\(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{16}{x^2}\) e.\(\left(1-\dfrac{x-1}{x+1}\right)\left(x+2\right)=\dfrac{x+1}{x-1}+\dfrac{x-1}{x+1}\)
bạn tách một câu vài câu hỏi chứ đừng gộp như thế này ko ai trả lời đâu
a: =>\(4x-5=2x-2+x=3x-2\)
=>x=3
b: \(\Leftrightarrow7x-35=3x+6\)
=>4x=41
=>x=41/4
c: =>(2x+5)(x+5)-2x^2=0
=>2x^2+10x+5x+25-2x^2=0
=>15x=-25
=>x=-5/3
e: \(\Leftrightarrow\dfrac{11}{x}=\dfrac{9x-36+2x+2}{\left(x+1\right)\left(x-4\right)}\)
=>11(x^2-3x-4)=x(11x-34)
=>11x^2-33x-44=11x^2-34x
=>x=44
tìm điều kiện bài toán:
a) \(y=\dfrac{1}{x}-\dfrac{\sqrt{2x-1}}{x^2-3x+2}\)
b) \(y=\dfrac{1}{x^2-1}-\sqrt{7-2x}\)
c) \(y=\dfrac{2}{x}+\dfrac{3}{4-2x+x^2}\)
d) \(y=\sqrt{25-x^2}-2\sqrt{x}+3\)
Lời giải:
a.
\(\left\{\begin{matrix} x\neq 0\\ 2x-1\geq 0\\ x^2-3x+2=(x-1)(x-2)\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\neq 0\\ x\geq \frac{1}{2}\\ x\neq 1; x\neq 2\end{matrix}\right.\)
$\Leftrightarrow x\geq \frac{1}{2}; x\neq 1; x\neq 2$
b. \(\left\{\begin{matrix}
x^2-1=(x-1)(x+1)\neq 0\\
7-2x\geq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix}
x\neq \pm 1\\
x\leq \frac{7}{2}\end{matrix}\right.\)
c.
\(\left\{\begin{matrix} x\neq 0\\ 4-2x+x^2\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\neq 0\\ (x-1)^2+3\neq 0\end{matrix}\right.\Leftrightarrow x\neq 0\)
d.
\(\left\{\begin{matrix} 25-x^2=(5-x)(5+x)\geq 0\\ x\geq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} -5\leq x\leq 5\\ x\geq 0\end{matrix}\right.\Leftrightarrow 0\leq x\leq 5\)
a) \(y=\dfrac{1}{x}-\dfrac{\sqrt[]{2x-1}}{x^2-3x+2}\)
Điều kiện \(\) \(2x-1\ge0;x\ne0;x^2-3x+2\ne0\)
\(\Leftrightarrow x\ge\dfrac{1}{2};x\ne0;\left(x-1\right)\left(x-2\right)\ne0\)
\(\Leftrightarrow x\ge\dfrac{1}{2};x\ne0;x\ne1;x\ne2\)
a) \(x\ge\dfrac{1}{2};x\ne1;x\ne2\)
b) \(x\le\dfrac{7}{2};x\ne\pm1\)
c) \(x\ne0\)
d) \(0\le x\le5\)
Tìm x, biết:
a) \(\sqrt{x^2-2x+1}=2\)
b)\(\sqrt{x^2-1}=x\)
c) \(\sqrt{4x-20}+3\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\)
d) \(x-5\sqrt{x-2}=-2\)
e) \(2x-3\sqrt{2x-1}-5=0\)
`a)sqrt{x^2-2x+1}=2`
`<=>sqrt{(x-1)^2}=2`
`<=>|x-1|=2`
`**x-1=2<=>x=3`
`**x-1=-1<=>x=-1`.
Vậy `S={3,-1}`
`b)sqrt{x^2-1}=x`
Điều kiện:\(\begin{cases}x^2-1 \ge 0\\x \ge 0\\\end{cases}\)
`<=>` \(\begin{cases}x^2 \ge 1\\x \ge 0\\\end{cases}\)
`<=>x>=1`
`pt<=>x^2-1=x^2`
`<=>-1=0` vô lý
Vậy pt vô nghiệm
`c)sqrt{4x-20}+3sqrt{(x-5)/9}-1/3sqrt{9x-45}=4(x>=5)`
`pt<=>sqrt{4(x-5)}+sqrt{9*(x-5)/9}-sqrt{(9x-45)*1/9}=4`
`<=>2sqrt{x-5}+sqrt{x-5}-sqrt{x-5}=4`
`<=>2sqrt{x-5}=4`
`<=>sqrt{x-5}=2`
`<=>x-5=4`
`<=>x=9(tmđk)`
Vậy `S={9}.`
`d)x-5sqrt{x-2}=-2(x>=2)`
`<=>x-2-5sqrt{x-2}+4=0`
Đặt `a=sqrt{x-2}`
`pt<=>a^2-5a+4=0`
`<=>a_1=1,a_2=4`
`<=>sqrt{x-2}=1,sqrt{x-2}=4`
`<=>x_1=3,x_2=18`,
`e)2x-3sqrt{2x-1}-5=0`
`<=>2x-1-3sqrt{2x-1}-4=0`
Đặt `a=sqrt{2x-1}(a>=0)`
`pt<=>a^2-3a-4=0`
`a-b+c=0`
`<=>a_1=-1(l),a_2=4(tm)`
`<=>sqrt{2x-1}=4`
`<=>2x-1=16`
`<=>x=17/2(tm)`
Vậy `S={17/2}`
d.
ĐKXĐ: $x\geq 2$. Đặt $\sqrt{x-2}=a(a\geq 0)$ thì pt trở thành:
$a^2+2-5a=-2$
$\Leftrightarrow a^2-5a+4=0$
$\Leftrightarrow (a-1)(a-4)=0$
$\Rightarrow a=1$ hoặc $a=4$
$\Leftrightarrow \sqrt{x-2}=1$ hoặc $\sqrt{x-2}=4$
$\Leftrightarrow x=3$ hoặc $x=18$ (đều thỏa mãn)
e. ĐKXĐ: $x\geq \frac{1}{2}$
Đặt $\sqrt{2x-1}=a(a\geq 0)$ thì pt trở thành:
$a^2+1-3a-5=0$
$\Leftrightarrow a^2-3a-4=0$
$\Leftrightarrow (a+1)(a-4)=0$
Vì $a\geq 0$ nên $a=4$
$\Leftrightarrow \sqrt{2x-1}=4$
$\Leftrightarrow x=\frac{17}{2}$
a.
$\sqrt{x^2-2x+1}=2$
$\Leftrightarrow \sqrt{(x-1)^2}=2$
$\Leftrightarrow |x-1|=2$
$\Rightarrow x-1=\pm 2$
$\Leftrightarrow x=3$ hoặc $x=-1$ (đều thỏa mãn)
b. ĐKXĐ: $x\geq 1$ hoặc $x\leq -1$
PT \(\Rightarrow \left\{\begin{matrix} x\geq 0\\ x^2-1=x^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 0\\ 1=0\end{matrix}\right.\) (vô lý)
Vậy pt vô nghiệm
c. ĐKXĐ: $x\geq 5$
PT $\Leftrightarrow \sqrt{4(x-5)}+3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9(x-5)}=4$
$\Leftrightarrow 2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4$
$\Leftrightarrow 2\sqrt{x-5}=4$
$\Leftrightarrow \sqrt{x-5}=2$
$\Leftrightarrow x=2^2+5=9$ (thỏa mãn)
Giải phương trình:
1. \(x^4-6x^2-12x-8=0\)
2. \(\dfrac{x}{2x^2+4x+1}+\dfrac{x}{2x^2-4x+1}=\dfrac{3}{5}\)
3. \(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)
4. \(2x^2.\sqrt{-4x^4+4x^2+3}=4x^4+1\)
5. \(x^2+4x+3=\sqrt{\dfrac{x}{8}+\dfrac{1}{2}}\)
6. \(\left\{{}\begin{matrix}4x^3+xy^2=3x-y\\4xy+y^2=2\end{matrix}\right.\)
7. \(\left\{{}\begin{matrix}\sqrt{x^2-3y}\left(2x+y+1\right)+2x+y-5=0\\5x^2+y^2+4xy-3y-5=0\end{matrix}\right.\)
8. \(\left\{{}\begin{matrix}\sqrt{2x^2+2}+\left(x^2+1\right)^2+2y-10=0\\\left(x^2+1\right)^2+x^2y\left(y-4\right)=0\end{matrix}\right.\)
1.
\(x^4-6x^2-12x-8=0\)
\(\Leftrightarrow x^4-2x^2+1-4x^2-12x-9=0\)
\(\Leftrightarrow\left(x^2-1\right)^2=\left(2x+3\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=2x+3\\x^2-1=-2x-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\x^2+2x+2=0\end{matrix}\right.\)
\(\Leftrightarrow x=1\pm\sqrt{5}\)
3.
ĐK: \(x\ge-9\)
\(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left(\sqrt{x+9}+x^2-9\right)=0\)
\(\Leftrightarrow\sqrt{x+9}+x^2-9=0\left(1\right)\)
Đặt \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)
\(\left(1\right)\Leftrightarrow t+x^2+x-t^2=0\)
\(\Leftrightarrow\left(x+t\right)\left(x-t+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-t\\x=t-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{x+9}\\x=\sqrt{x+9}-1\end{matrix}\right.\)
\(\Leftrightarrow...\)
2.
ĐK: \(x\ne\dfrac{2\pm\sqrt{2}}{2};x\ne\dfrac{-2\pm\sqrt{2}}{2}\)
\(\dfrac{x}{2x^2+4x+1}+\dfrac{x}{2x^2-4x+1}=\dfrac{3}{5}\)
\(\Leftrightarrow\dfrac{1}{2x+\dfrac{1}{x}+4}+\dfrac{1}{2x+\dfrac{1}{x}-4}=\dfrac{3}{5}\)
Đặt \(2x+\dfrac{1}{x}+4=a;2x+\dfrac{1}{x}-4=b\left(a,b\ne0\right)\)
\(pt\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{3}{5}\left(1\right)\)
Lại có \(a-b=8\Rightarrow a=b+8\), khi đó:
\(\left(1\right)\Leftrightarrow\dfrac{1}{b+8}+\dfrac{1}{b}=\dfrac{3}{5}\)
\(\Leftrightarrow\dfrac{2b+8}{\left(b+8\right)b}=\dfrac{3}{5}\)
\(\Leftrightarrow10b+40=3\left(b+8\right)b\)
\(\Leftrightarrow\left[{}\begin{matrix}b=2\\b=-\dfrac{20}{3}\end{matrix}\right.\)
TH1: \(b=2\Leftrightarrow...\)
TH2: \(b=-\dfrac{20}{3}\Leftrightarrow...\)
a,\(\dfrac{x+1}{x-3}+\dfrac{-2x^2+2x}{x^2-9}+\dfrac{x-1}{x+3}\)
b,\(\dfrac{1-2x}{6x^3y}+\dfrac{3+2y}{6x^3y}+\dfrac{2x-4}{6x^3y}\)
c,\(\dfrac{5}{2x^2y}+\dfrac{3}{5xy^2}+\dfrac{x}{3y^3}\)
d,\(\dfrac{5}{4\left(x+2\right)}+\dfrac{8-x}{4x^2+8x}\)
c,\(\dfrac{x^2+2}{x^3+1}+\dfrac{2}{x^2+x+1}+\dfrac{1}{1-x}\)
\(a,=\dfrac{x^2+4x+3-2x^2+2x+x^2-4x+3}{\left(x-3\right)\left(x+3\right)}=\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x-3}\\ b,=\dfrac{1-2x+3+2y+2x-4}{6x^3y}=\dfrac{2y}{6x^3y}=\dfrac{1}{x^2}\\ c,=\dfrac{75y^2+18xy+10x^2}{30x^2y^3}\\ d,=\dfrac{5x+8-x}{4x\left(x+2\right)}=\dfrac{4\left(x+2\right)}{4x\left(x+2\right)}=\dfrac{1}{x}\\ c,=\dfrac{x^2+2+2x-2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1}{x^2+x+1}\)
1)Tìm x để căn thức sau có nghĩa
a)\(\sqrt{2x-4}\) b)\(\sqrt{\dfrac{-7}{4-x}}\)
2) Tính
A=\(\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}
\)
B=\(\left(\dfrac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\dfrac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}\)
Helpppp
1)
a) \(\sqrt{2x-4}\) có nghĩa khi:
\(2x-4\ge0\)
\(\Leftrightarrow2x\ge4\)
\(\Leftrightarrow x\ge\dfrac{4}{2}\)
\(\Leftrightarrow x\ge2\)
b) \(\sqrt{\dfrac{-7}{4-x}}\) có nghĩa khi
\(\dfrac{-7}{4-x}\ge0\) mà \(-7< 0\)
\(\Rightarrow4-x\le0\)
\(\Leftrightarrow x\ge4\)
2)
a) \(A=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(A=\sqrt{\left(\sqrt{5}\right)^2+2\cdot2\sqrt{5}+2^2}-\sqrt{\left(\sqrt{5}\right)^2-2\cdot2\cdot\sqrt{5}+2^2}\)
\(A=\sqrt{\left(\sqrt{5}+2\right)^2}-\sqrt{\left(\sqrt{5}-2\right)^2}\)
\(A=\left|\sqrt{5}+2\right|-\left|\sqrt{5}-2\right|\)
\(A=\sqrt{5}+2-\sqrt{5}+2\)
\(A=4\)
\(B=\left(\dfrac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\dfrac{\sqrt{15}-5}{1-\sqrt{3}}\right):\dfrac{1}{\sqrt{5}-\sqrt{7}}\)
\(B=\left(-\dfrac{\sqrt{14}-\sqrt{7}}{\sqrt{2}-1}-\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}\)
\(B=\left[-\dfrac{\sqrt{7}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}-\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\right]\cdot\left(\sqrt{7}-\sqrt{5}\right)\)
\(B=\left(-\sqrt{7}-\sqrt{5}\right)\cdot\left(\sqrt{7}+\sqrt{5}\right)\)
\(B=-\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)\)
\(B=-\left(7-5\right)\)
\(B=-2\)
Tìm điều kiện để các biểu thức sau xác định
a)\(\sqrt{x+1}-\dfrac{1}{2}\)
b)\(2.\sqrt{1-2x}-\dfrac{\sqrt{3}-1}{4}\)
c)\(\sqrt{x+1}-\sqrt{x-2}\)
d)\(\sqrt{2-3x}-\sqrt{1-2x}\)
e)\(2.\sqrt{\sqrt{3}-2x}+\dfrac{1}{x-1}\)
f)\(\dfrac{1}{2}.\sqrt{x-\dfrac{\sqrt{3}}{2}}-\dfrac{1}{\sqrt{x}-1}\)
g)\(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}+2}\)
h)\(\dfrac{1}{\sqrt{x}+1}-\dfrac{1}{\sqrt{x^2+2}}\)
a, \(x+1\ge0\Leftrightarrow x\ge-1\)
b, \(1-2x\ge0\Leftrightarrow x\le\dfrac{1}{2}\)
c, \(\left\{{}\begin{matrix}x+1\ge0\\x-2\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x\ge2\end{matrix}\right.\Leftrightarrow x\ge2\)
d, \(\left\{{}\begin{matrix}2-3x\ge0\\1-2x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{2}{3}\\x\le\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow x\le\dfrac{1}{2}\)
e, \(\left\{{}\begin{matrix}\sqrt{3}-2x\ge0\\x-1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{\sqrt{3}}{2}\\x\ne1\end{matrix}\right.\Leftrightarrow x\le\dfrac{\sqrt{3}}{2}\)
f, \(\left\{{}\begin{matrix}x-\dfrac{\sqrt{3}}{2}\ge0\\x\ge0\\\sqrt{x}-1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{\sqrt{3}}{2}\\x\ge0\\x\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{\sqrt{3}}{2}\\x\ne1\end{matrix}\right.\)
g, \(\left\{{}\begin{matrix}\sqrt{x}-1\ne0\\\sqrt{x}+2\ne0\\x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)