Tìm x bt:
a, /x/ = 1/2
b, /x/ = 3,12
c, /x/ = 0
/x/ = 2 \(^{\frac{1}{7}}\)
1.cho bt:A=\(\frac{x^3}{x^2-4}-\frac{x}{x-2}-\frac{2}{x+2}\)
a) Rút gọn A
b)Tìm x để A>0,A<0
c)Tìm X để /A/=5
ĐK : \(x\ne2\); \(x\ne-2\)
a) \(A=\frac{x^3}{x^2-4}-\frac{x}{x-2}-\frac{2}{x+2}=\frac{x^3}{\left(x-2\right)\left(x+2\right)}-\frac{x}{x-2}-\frac{2}{x+2}\)
\(=\frac{x^3-x.\left(x+2\right)-2.\left(x-2\right)}{\left(x+2\right).\left(x-2\right)}=\frac{x^3-x^2-2x-2x+4}{\left(x+2\right).\left(x-2\right)}=\frac{x^3-x^2-4x+4}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{x^2.\left(x-1\right)-4.\left(x-1\right)}{\left(x+2\right)\left(x-2\right)}=\frac{\left(x-1\right).\left(x^2-4\right)}{\left(x+2\right)\left(x-2\right)}=\frac{\left(x-1\right)\left(x+2\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=x-1\)
b) - Để A > 0 thì x - 1 > 0 => x > 1
- Để A < 0 thì x - 1 < 0 => x < 1
c) Để | A | = 5 thì | x-1 | = 5
+ Nếu \(x-1\ge0\) thì \(x\ge1\) , ta có phương trình
x - 1 = 5 => x = 6 ( thỏa mãn )
+ Nếu x - 1 < 0 thì x < 1 , ta có phương trình :
-x + 1 = 5 < = > -x = 4 <=> x = -4 ( thỏa mãn )
Vậy tập nghiệm của phương trình là S = { -4 ; 6 }
1/ cho bt:A=\(\frac{1-\sqrt{x}}{\sqrt{x}}+\frac{\sqrt{x}+4}{\sqrt{x}+1}-\frac{1}{x+\sqrt{x}}\)
tìm gt của x để P=\(\sqrt{x}\)+ A có gt nhỏ nhất
2/giải hệ pt:\(\hept{\begin{cases}xy+x+y-1=0\\yz+y+z-5=0\\zx+x+z-2=0\end{cases}}\)
1 Cho bt:A=\((\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}):\frac{\sqrt{x}+1}{x-2\sqrt{x}+1}\)
Rút gọn A
Tìm x để A>-1
ĐK: \(x>0;x\ne1\)
\(A=\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{x-2\sqrt{x}+1}\)
\(=\left(\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}}\)
\(A>-1\) \(\Rightarrow\)\(\frac{\sqrt{x}-1}{\sqrt{x}}>-1\)
\(\Leftrightarrow\)\(\frac{\sqrt{x}-1}{\sqrt{x}}+1>0\) \(\Leftrightarrow\)\(\frac{2\sqrt{x}-1}{\sqrt{x}}>0\)
Do \(\sqrt{x}>0\) \(\Rightarrow\)\(2\sqrt{x}-1>0\)\(\Leftrightarrow\)\(2\sqrt{x}>1\)\(\Leftrightarrow\)\(\sqrt{x}>\frac{1}{2}\)\(\Leftrightarrow\)\(x>\frac{1}{4}\)
Vậy \(x>\frac{1}{4}\)\(\left(x\ne1\right)\)thì A > - 1
\(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne1\end{cases}}\)
Ta có: \(A=\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{x-2\sqrt{x}+1}\)\(=\left[\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{1}{\sqrt{x}-1}\right]:\frac{\sqrt{x}+1}{\left(\sqrt{x}\right)^2-2\sqrt{x}+1}\)
\(=\left[\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right]:\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}=\frac{\sqrt{x}-1}{\sqrt{x}}\)
Để \(A>-1\)thì \(\frac{\sqrt{x}-1}{\sqrt{x}}>-1\)\(\Leftrightarrow\sqrt{x}-1>-\sqrt{x}\)\(\Leftrightarrow2\sqrt{x}>1\)
\(\Leftrightarrow\sqrt{x}>\frac{1}{2}\)\(\Leftrightarrow x>\frac{1}{4}\)thoả mãn \(x\ne1\)
Vậy \(A>-1\)\(\Leftrightarrow x>\frac{1}{4}\)thoả mãn \(x\ne1\)
cho bt:A=(1/x-2-2x/4-x^2+1/2+x)*(2/x-1)
câu hỏi:rút gọn và tính giá trị bt tại x thỏa mãn 2x^2+x=0
tìm x để A=1/2
tìm x nguyên để A nguyên dương
1 Cho bt:A=\((\frac{1}{x+\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x+1}}):\frac{3}{\sqrt{x+1}}\)
Đkxđ và rút gọn
Tìm x để \(\frac{1}{\sqrt{x}}.A=1\)
* Tìm x, bt:
a.\(\sqrt{\left(2x+3\right)^2}=8\)
b.\(\sqrt{9x}-7\sqrt{x}=8-6\sqrt{x}\)
c.\(\sqrt{9x-9}+1=13\)
a) ⇔ |2x+3| = 8
⇒ \(\left[{}\begin{matrix}2x+3=8\\2x+3=-8\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}2x=5\\2x=-11\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{11}{2}\end{matrix}\right.\)
Vậy...
b) ĐKXĐ: \(x\ge0\)
\(\Leftrightarrow3\sqrt{x}-7\sqrt{x}+6\sqrt{x}=8\)
\(\Leftrightarrow2\sqrt{x}=8\)
\(\Leftrightarrow\sqrt{x}=4\)
\(\Leftrightarrow x=16\) (Vì \(x\ge0\) )
Vậy x = 16
c) ĐKXĐ: \(x\ge1\)
\(\Leftrightarrow\sqrt{9\left(x-1\right)}=12\)
\(\Leftrightarrow3\sqrt{x-1}=12\)
\(\Leftrightarrow\sqrt{x-1}=4\)
\(\Leftrightarrow x-1=16\)
\(\Leftrightarrow x=17\)(TM)
Vậy x = 17
rút gọn bt:
a,(x+y)^2+(x-y)^2
b,(a-b^2)(a+b^2)
a) `(x+y)^2+(x-y)^2=x^2+2xy+y^2+x^2-2xy+y^2=2x^2+2y^2`
b) `(a-b^2)(a+b^2)=a^2-(b^2)^2=a^2-b^4`
1 Cho bt:A=\((\frac{1}{x+\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}+1}):\frac{3}{\sqrt{x}+1}\)
Đkxđ và rút gọn
A = \(\frac{1+x}{x+\sqrt{x}}.\frac{\sqrt{x}+1}{3}\)=\(\frac{1+x}{3\sqrt{x}}\)
ĐKXĐ : x > 0
Tìm GTNN bt:A=\(\dfrac{2020x+2021\sqrt{1-x^2}+2022}{\sqrt{1-x^2}}\)
1. Cho a,b \(\ge\)0 và a+b\(\le\)2 . Chứng minh \(\frac{2+\alpha}{1+\alpha}+\frac{1-2b}{1+2b}\ge\frac{8}{7}\)
2. Tìm x: \(\frac{\frac{2x-1}{2}-3}{4}-\frac{4-\frac{1+2x}{3}}{2}=\frac{5-\frac{1}{2}X}{6}-3\)
3.tìm x: \(\left(x-2\right)^2-3\left(x-1\right)+2x^2\left(x-1\right)=\left(2x+1\right)^2+2\left(x^3-2\right)\)
4. Rút Gọn B = \(\left(\frac{X+2}{X+1}-\frac{2X}{X-1}\right)\div\frac{X}{3X+3}+\frac{4X^2+X+7}{X^2-X}\)