tìm x biết : \(x-2x+2^2\cdot x-2^3\cdot x+2^4\cdot x-....+2^{2006}\cdot x-2^{2007}\cdot x=2^{2008}-1\)
các bn giúp mk nhoa mk sẽ tick cho ai giải nhanh ; đúng và rõ ràng nhất
THANH YOU NHIỀU
bài 1: tìm x, biết
\(\frac{1}{4}\cdot\frac{2}{6}\cdot\frac{3}{8}\cdot\frac{4}{10}\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\frac{30}{62}\cdot\frac{31}{64}=2^x\)
bài 2:
cho: p = \(\left(x-4\right)^{\left(x-5\right)^{\left(x-6\right)^{\left(x+5\right)}}}\)
tính p(x)=7
giúp mk vs!!!!!
mk cần gấp!!
Bài làm:
Ta có: \(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.....\frac{30}{62}.\frac{31}{64}=2^x\)
\(\Leftrightarrow\frac{1.2.3.....30.31}{2.2.2.3.2.4.....2.31.2.32}=2^x\)
\(\Leftrightarrow\frac{1}{2^{31}.2^5}=2^x\)
\(\Leftrightarrow\frac{1}{2^{36}}=2^x\)
\(\Rightarrow x=-36\)
mk cần cả giải thích
giúp mk vs!!!
giải hệ phương trình :
a) \(\hept{\begin{cases}x\cdot\left(1+y-x\right)=-2\cdot y^2-y\\x\cdot\left(\sqrt{2\cdot y}-2\right)=y\cdot\left(\sqrt{x-1}-2\right)\end{cases}}\)
b) \(\hept{\begin{cases}1+x\cdot y+\sqrt{x\cdot y}=x\\\frac{1}{x\cdot\sqrt{x}}+y\cdot\sqrt{y}=\frac{1}{\sqrt{x}}+3\cdot\sqrt{y}\end{cases}}\)
Làm hộ mk nhé mk tick cho :))))))))))
giải pt
1/ \(x^4+2008\cdot x^2+2007\cdot x+2008=0\)
2/\(x^4+x^2-6x+8=0\)
1: \(\Leftrightarrow x^4+x^3+x^2-x^3-x^2-x+2008x^2+2008x+2008=0\)
\(\Leftrightarrow\left(x^2+x+1\right)\left(x^2-x+2008\right)=0\)
hay \(x\in\varnothing\)
2: \(x^4+x^2+6x-8=0\)
\(\Leftrightarrow x^4-x^3+x^3-x^2+2x^2-2x+8x-8=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+x^2+2x+8\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2-x+4\right)=0\)
hay \(x\in\left\{1;-2\right\}\)
Tìm x :
\(3x\cdot\left(x-2\right)-2x\cdot\left(2x-1\right)=\left(1-x\right)\cdot\left(1+x\right)\)
\(\left(5x+3\right)\cdot\left(3x-5\right)-\left(x-2\right)\cdot\left(2x+1\right)=6x\cdot\left(3x+1\right)-x^2\)
\(\left(2x-1\right)\cdot\left(2x+1\right)-3\cdot\left(x-1\right)=\left(1-4x\right)\cdot\left(1-x\right)\)
\(\left(2x^2+1\right)\cdot\left(3x^2-1\right)-\left(4x^2-3\right)\cdot\left(x^2+1\right)=x\cdot\left(2x^3+1\right)\)
GIÚP MK ĐI MAI MK PHẢI NỘP RÙI !
1> 3x(x-2)-2x(2x-1)=(1-x)(1+x)
⇔\(3x^2\)-6x-\(4x^2\)+2x=1-\(x^2\)
⇔-1\(x^2\) - 4x= 1- \(x^2\)
⇔ -1\(x^2\) -4x+ \(x^2\) = 1
⇔-4x=1
⇔ x = \(\dfrac{-1}{4}\)
1,Tìm x, biết:
a/\(^{\left(x-1\right)^{x+2}}=^{\left(x-1\right)^{x+4}}\)
b/\(\dfrac{1}{4}\cdot\dfrac{2}{6}\cdot\dfrac{3}{8}\cdot\dfrac{4}{10}\cdot...\cdot\dfrac{30}{62}\cdot\dfrac{31}{64}=2x\)
c/\(4^x+3^x=2^x+6^x\left(STN\right)\)
GIÚP MÌNH NHANH NHA MÌNH TICK CHO
a)x=1;2;-2(bạn nên tự giải)
b)=>\(\dfrac{1\cdot2\cdot3\cdot4\cdot...\cdot30\cdot31}{4\cdot6\cdot8\cdot10\cdot...\cdot62\cdot64}\)=2x
=>\(\dfrac{2\cdot3\cdot4\cdot5\cdot...\cdot30\cdot31}{60\left(2\cdot3\cdot4\cdot5\cdot...\cdot30\cdot31\right)\cdot64}=2x\)
=>\(\dfrac{1}{60\cdot64}=2x\)=> 1/3840 =2x
=>x = 1/7680
c)=>4x - 2x = 6x - 3x
=>2x (2x-1)= 3x(2x-1)
=> 2x = 3x
=>x = 0
tìm x
a) \(\frac{x-1}{2}+\frac{x-2}{5}=\frac{1}{4}+\frac{x-7}{10}\)
b) \(3-\frac{2}{2x-3}=\frac{2}{5}+\frac{1}{2x-3}-\frac{3}{2}\)
c)\(7\cdot\left(x-1\right)+2x\cdot\left(1-x\right)=0\)
d) \(\frac{x+1}{2008}+\frac{x+2}{2017}+\frac{x+3}{2016}=\frac{x+10}{2009}+\frac{x+11}{2008}+\frac{x+12}{2007}\)
e) \(\frac{2}{\left(x-1\right)\cdot\left(x-3\right)}+\frac{5}{\left(x-3\right)\cdot\left(x-8\right)}+\frac{12}{\left(x-8\right)\cdot\left(x-20\right)}-\frac{1}{x-20}=\frac{-3}{4}\)
1, Giải hệ phương trình:
\(\hept{\begin{cases}x\cdot\left|x\right|-\left(x+10\right)\cdot\left|x+10\right|=y\cdot\left|y\right|\\y\cdot\left|y\right|-\left(y+10\right)\cdot\left|y+10\right|=z\cdot\left|z\right|\\z\cdot\left|z\right|-\left(z+10\right)\cdot\left|z+10\right|=x\cdot\left|x\right|\end{cases}}\)
Giải hộ mk nhoa mk tick cho !!!!!!!!!
Giải các phương trình sau
a) \(-x^2+4\cdot x+1=2\cdot\sqrt{2\cdot x+1}\)
b) \(x+\sqrt{x+\dfrac{1}{2}+\sqrt{x+\dfrac{1}{4}}}=2\)
c) \(5\cdot x^2-2\cdot x+1=\left(4\cdot x-1\right)\cdot\sqrt{x^2+1}\)
d) \(\left(2\cdot x-1\right)\cdot\sqrt{10-4\cdot x^2}=5-2\cdot x\)
e) \(\sqrt{2\cdot x-1}-\sqrt{x+1}=2\cdot x-4\)
f) \(\sqrt{x^2-2\cdot x}+\sqrt{2\cdot x^2+4\cdot x}=2\cdot x\)
câu b đk x>= -1/4
\(x+\sqrt{x+\dfrac{1}{2}+\sqrt{x+\dfrac{1}{4}}}=2\)
\(x+\sqrt{\left(\sqrt{x+\dfrac{1}{4}}+\dfrac{1}{2}\right)^2}=2\)
\(\left(\sqrt{x+\dfrac{1}{4}}+\dfrac{1}{2}\right)^2=2\)
\(x+\dfrac{1}{4}=\left(\sqrt{2}-\dfrac{1}{2}\right)^2\)
\(x=\left(\sqrt{2}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\)
\(x=\left(\sqrt{2}-\dfrac{1}{2}-\dfrac{1}{2}\right)\left(\sqrt{2}-\dfrac{1}{2}+\dfrac{1}{2}\right)\)
\(x=\sqrt{2}\left(\sqrt{2}-1\right)=2-\sqrt{2}\)
Tìm x,y,z biết :
a)\(\frac{3}{4}\cdot\left(x+\frac{1}{2}\right)=\frac{1}{2}\)
b) \(x\cdot\left(x-\frac{3}{2}\right)=0\)
c)\(|\frac{1}{4}-x|+\frac{2}{3}=\frac{1}{2}\)
d)\(|3\cdot x-2|=|x+1|\)
e)\(|x+\frac{3}{4}|+|x-y|=0\)
g)\(|4\cdot x-2|-2\cdot x=\frac{3}{4}\)
h)\(|5\cdot x+1|-2\cdot x=1\)
i)\((x-\frac{3}{2})\cdot\left(2\cdot x+1\right)>0\)
k)\(|x+\frac{1}{2}|+|x+y+z|+|\frac{1}{3}+y|=0\)
l)\(|2\cdot x-\frac{1}{2}|-\frac{1}{4}=3\)
GIÚP MK VS, CẢM ƠN MỌI NGƯỜI RẤT NHÌU !!!
lam sao bạn viết chữ to như 3/4 . ( x + 1/2 ) vậy