3) tìm x:
a) \(x^2=\dfrac{1}{16}\)
b)\(x^5\div x^2=\dfrac{-1}{64}\)
c)\(x^3\div x^2=\dfrac{32}{243}\)
d)\(\left(x^2\right)^2=\dfrac{81}{16}\)
Tìm x:
\(a\)) \(\dfrac{2}{3}+\left(x-\dfrac{1}{2}\right)^3=\dfrac{19}{27}\)
\(b\)) \(\left(\dfrac{3}{2}\right)^{2x-1}:\left(\dfrac{27}{8}\right)^3=\dfrac{81}{16}\)
\(c\)) \(\dfrac{1}{2}.2^x+4.2^x=9.2^5\)
\(d\)) \(\text{12 - (2x +1)}^2=-69\)
\(a,\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}=\left(\dfrac{1}{3}\right)^3\\ \Rightarrow x-\dfrac{1}{2}=\dfrac{1}{3}\Rightarrow x=\dfrac{5}{6}\\ b,\Rightarrow\left(\dfrac{3}{2}\right)^{2x-1}:\left(\dfrac{3}{2}\right)^9=\left(\dfrac{3}{2}\right)^4\\ \Rightarrow2x-1-9=4\\ \Rightarrow2x=14\Rightarrow x=7\\ c,\Rightarrow2^{x-1}+2^{x+2}=9\cdot2^5\\ \Rightarrow2^{x-1}\left(1+2^3\right)=9\cdot2^5\\ \Rightarrow2^{x-1}\cdot9=9\cdot2^5\\ \Rightarrow2^{x-1}=2^5\Rightarrow x-1=5\Rightarrow x=6\\ d,\Rightarrow\left(2x+1\right)^2=12+69=81\\ \Rightarrow\left[{}\begin{matrix}2x+1=9\\2x+1=-9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-5\end{matrix}\right.\)
\(a,\dfrac{2}{3}+\left(x-\dfrac{1}{2}\right)^3=\dfrac{19}{27}\)
\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{19}{27}-\dfrac{2}{3}\)
\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\left(\dfrac{1}{3}\right)^3\)
\(\Rightarrow x-\dfrac{1}{2}=\dfrac{1}{3}\)
\(x=\dfrac{1}{2}+\dfrac{1}{3}\)
\(x=\dfrac{1}{5}\)
\(A=\left(\dfrac{4x}{x+2}-\dfrac{x^3-8}{x^3+8}\times\dfrac{4x^2-8x+16}{x^2-4}\right)\div\dfrac{16}{x+2}\times\dfrac{x^2+3x+2}{x^2+x+1}\)
\(B=\dfrac{x^2+x-2}{x^3-1}\)
a) Tìm ĐKXĐ của A, B. Rút gọn A, B
b)Tìm GTLN của A+B
Tìm x biết:
\(a,\left(x-\dfrac{3}{4}\right)+50\%=\dfrac{1}{6}\)
\(b,\dfrac{1}{2}x-\dfrac{5}{6}x=\dfrac{7}{2}\)
\(c,\left(4-x\right)\left(3x+5\right)=0\)
\(d,\dfrac{x}{16}=\dfrac{50}{32}\)
\(e,\left(2x-3\right)+\dfrac{3}{2}=-\dfrac{1}{4}\)
a: =>x-3/4=1/6-1/2=1/6-3/6=-2/6=-1/3
=>x=-1/3+3/4=-4/12+9/12=5/12
b: =>x(1/2-5/6)=7/2
=>-1/3x=7/2
hay x=-21/2
c: (4-x)(3x+5)=0
=>4-x=0 hoặc 3x+5=0
=>x=4 hoặc x=-5/3
d: x/16=50/32
=>x/16=25/16
hay x=25
e: =>2x-3=-1/4-3/2=-1/4-6/4=-7/4
=>2x=-7/4+3=5/4
hay x=5/8
giải các phương trình sau
a) \(2^{x^2-1}=256\)
b) \(3^{x^2+3x}=81\)
c) \(2^{x^2-5x}=64\)
d) \(\left(\dfrac{1}{3}\right)^x=243\)
e) \(\left(\dfrac{1}{3}\right)^{x+5}=3^{2x+1}\)
a: \(2^{x^2-1}=256\)
=>\(2^{x^2-1}=2^8\)
=>\(x^2-1=8\)
=>\(x^2=9\)
=>\(x\in\left\{3;-3\right\}\)
b: \(3^{x^2+3x}=81\)
=>\(3^{x^2+3x}=3^4\)
=>\(x^2+3x=4\)
=>\(x^2+3x-4=0\)
=>(x+4)(x-1)=0
=>\(\left[{}\begin{matrix}x+4=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=1\end{matrix}\right.\)
c: \(2^{x^2-5x}=64\)
=>\(2^{x^2-5x}=2^6\)
=>\(x^2-5x=6\)
=>\(x^2-5x-6=0\)
=>(x-6)(x+1)=0
=>\(\left[{}\begin{matrix}x-6=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-1\end{matrix}\right.\)
d: \(\left(\dfrac{1}{3}\right)^x=243\)
=>\(\left(\dfrac{1}{3}\right)^x=3^5=\left(\dfrac{1}{3}\right)^{-5}\)
=>x=-5
e: \(\left(\dfrac{1}{3}\right)^{x+5}=3^{2x+1}\)
=>\(3^{-x-5}=3^{2x+1}\)
=>-x-5=2x+1
=>-3x=6
=>x=-2
Thực hiện phép tính:
a) \(\dfrac{x+2y}{xy}\div\dfrac{x^2+4xy+4y^2}{2x^2}\)
b) \(\dfrac{4x^3-xy^2}{x^2+xy+y^2}\div\dfrac{\left(2x-y\right)^3}{x^3-y^3}\)
c) \(\dfrac{x+3}{x+2}\div\dfrac{3x+9}{2x-1}\div\dfrac{4x-2}{2x+4}\)
d) \(\dfrac{x+1}{x+2}\div\left(\dfrac{2x^2}{2x-3}\times\dfrac{3x+3}{4x^3}\right)\)
a: \(=\dfrac{x+2y}{xy}\cdot\dfrac{2x^2}{\left(x+2y\right)^2}=\dfrac{2x}{y\left(x+2y\right)}\)
b: \(=\dfrac{x\left(4x^2-y^2\right)}{x^2+xy+y^2}\cdot\dfrac{\left(x-y\right)\left(x^2+xy+y^2\right)}{\left(2x-y\right)^3}\)
\(=\dfrac{x\left(x-y\right)\left(2x+y\right)\left(2x-y\right)}{\left(2x-y\right)^3}\)
\(=\dfrac{x\left(x-y\right)\left(2x+y\right)}{\left(2x-y\right)^2}\)
c: \(=\dfrac{x+3}{x+2}\cdot\dfrac{2x-1}{3\left(x+3\right)}\cdot\dfrac{2\left(x+2\right)}{2\left(2x-1\right)}\)
=1/3
d: \(=\dfrac{x+1}{x+2}:\left(\dfrac{1}{2x}\cdot\dfrac{3x+3}{2x-3}\right)\)
\(=\dfrac{x+1}{x+2}\cdot\dfrac{2x\left(2x-3\right)}{3\left(x+1\right)}=\dfrac{2x\left(2x-3\right)}{3\left(x+2\right)}\)
Tìm x:
a) \(\dfrac{1}{3}.x+\dfrac{2}{5}\left(x-1\right)=0\)
b)\(-5.\left(x+\dfrac{1}{5}\right)-\dfrac{1}{2}.\left(x-\dfrac{2}{3}\right)=x\)
c)\(\left(x+\dfrac{1}{2}\right).\left(\dfrac{2}{3}-2x\right)=0\)
d)\(9.\left(3x+1\right)^2=16\)
a: =>1/3x+2/5x-2/5=0
=>11/15x-2/5=0
=>11/15x=2/5
=>x=2/5:11/15=2/5*15/11=30/55=6/11
b: =>-5x-1-1/2x+1/3=x
=>-11/2x-2/3-x=0
=>-13/2x=2/3
=>x=-2/3:13/2=-2/3*2/13=-4/39
c: (x+1/2)(2/3-2x)=0
=>x+1/2=0 hoặc 2/3-2x=0
=>x=1/3 hoặc x=-1/2
d: 9(3x+1)^2=16
=>(3x+1)^2=16/9
=>3x+1=4/3 hoặc 3x+1=-4/3
=>3x=1/3 hoặc 3x=-7/3
=>x=1/9 hoặc x=-7/9
\(C=\left(\dfrac{2\sqrt{x}}{2x-5\sqrt{x}+3}-\dfrac{5}{2\sqrt{x}-3}\right)\div\left(3+\dfrac{2}{1-\sqrt{x}}\right)\)
a) Rút gọn C
b) Tính C với \(x=\dfrac{2}{2-\sqrt{3}}\)
c) Tìm x để C= –1
d) Tìm x để C > 0
e) So sánh C’ với –2
f) Tìm GRNN của C’ với C’=\(\dfrac{1}{C}\times\dfrac{1}{\sqrt{x}+1}\)
i)Tìm \(x\in Z\) để \(C'\in Z\) g) Tìm m để pt C’.m = –1 có nghiệm
ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\notin\left\{1;\dfrac{25}{9};\dfrac{9}{4}\right\}\end{matrix}\right.\)
a: \(C=\left(\dfrac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(2\sqrt{x}-3\right)}-\dfrac{5}{2\sqrt{x}-3}\right):\left(3-\dfrac{2}{\sqrt{x}-1}\right)\)
\(=\dfrac{2\sqrt{x}-5\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(2\sqrt{x}-3\right)}:\dfrac{3\sqrt{x}-3-2}{\sqrt{x}-1}\)
\(=\dfrac{2\sqrt{x}-5\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(2\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-1}{3\sqrt{x}-5}\)
\(=-\dfrac{1}{2\sqrt{x}-3}\)
b: \(x=\dfrac{2}{2-\sqrt{3}}=2\left(2+\sqrt{3}\right)=4+2\sqrt{3}\)
Khi \(x=4+2\sqrt{3}\) thì \(C=-\dfrac{1}{2\left(\sqrt{3}+1\right)-3}=\dfrac{-1}{2\sqrt{3}-1}=\dfrac{-2\sqrt{3}-1}{11}\)
c: C=-1
=>\(2\sqrt{x}-3=1\)
=>\(\sqrt{x}=2\)
=>x=4(nhận)
d: C>0
=>\(2\sqrt{x}-3< 0\)
=>\(\sqrt{x}< \dfrac{3}{2}\)
=>\(0< =x< \dfrac{9}{4}\)
Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0< =x< \dfrac{9}{4}\\x< >1\end{matrix}\right.\)
cho B=\(\left(\dfrac{2\sqrt{x}+x}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right)\div\left(1-\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\)
a. rút gọn B
b. tính \(\sqrt{B}\) khi \(x=5+2\sqrt{3}\)
c. tìm x để B= \(\dfrac{1}{2x^3-x-1}\)
d. tìm giá trị của x để giá trị của B không lớn hơn giá trị biểu thức \(\dfrac{1}{x+2}\)
Lm nhanh giúp mk nhé mk đang cần gấp
a) \(B=\left(\dfrac{2\sqrt{x}+x}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right):\left(1-\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\left(x\ge0,x\ne1\right)\)
\(=\left(\dfrac{2\sqrt{x}+x}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{1}{\sqrt{x}-1}\right):\dfrac{x+\sqrt{x}+1-\sqrt{x}-2}{x+\sqrt{x}+1}\)
\(=\dfrac{2\sqrt{x}+x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\dfrac{x-1}{x+\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{x+\sqrt{x}+1}{x-1}=\dfrac{1}{x-1}\)
a) Ta có: \(B=\left(\dfrac{2\sqrt{x}+x}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right):\left(1-\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\)
\(=\dfrac{x+2\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\dfrac{x+\sqrt{x}+1-\sqrt{x}-2}{\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{1}{x+\sqrt{x}+1}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)
\(=\dfrac{1}{x-1}\)
Tìm x ∈ N biết :
a) \(8< 2^x\le2^9.2^{-5}\)
b)\(27< 81^3:3^x< 243\)
c)\(\left(\dfrac{2}{5}\right)^x>\left(\dfrac{5}{2}\right)^{-3}.\left(\dfrac{-3}{5}\right)^2\)
\(a,\Rightarrow2^3< 2^x\le2^4\Rightarrow x=4\\ b,\Rightarrow3^3< 3^{12}:3^x< 3^5\\ \Rightarrow3^3< 3^{12-x}< 3^5\\ \Rightarrow12-x=4\Rightarrow x=8\)