Tìm x biết
1) \(\left|\dfrac{1}{4}\times x^2+\dfrac{1}{45}\right|+\dfrac{1}{5}=\dfrac{1}{4}\)
2) \(\left(x^2-3\right)\times x^2-2\times x\times\left(x^2-3\right)\)
Tìm x
a, \(\left(2-x\right)\times\left(\dfrac{4}{5}-x\right)< 0\)
b, \(\left(x-\dfrac{3}{2}\right)\times\left(2\times x-1\right)>0\)
c, \(\left(x-\dfrac{4}{7}\right)\div\left(x+\dfrac{1}{2}\right)>0\)
d, \(\left(2\times x-5\right)\div\left(x+1\dfrac{3}{4}\right)< 0\)
Giúp mk với
\(a,\left(2-x\right)\left(\dfrac{4}{5}-x\right)< 0\)
=>Trong 2 số phải có 1 số âm và 1 số dương
Mà \(2-x>\dfrac{4}{5}-x\)
=>\(\dfrac{4}{5}< x< 2\)
Vậy...
Bài 2: Tìm x:
a)\(\dfrac{x-1}{27}\)=\(\dfrac{-3}{1-x}\) c)\(3\times x=2\times y\) và\(x-2\times y=8\)
b)\(\dfrac{4}{5}\)-\(\left|x-\dfrac{1}{2}\right|\)=\(\dfrac{3}{4}\) d)\(\dfrac{x-1}{2005}\)=\(\dfrac{3-y}{2006}\) và x-4009=y
a: \(\Leftrightarrow\left(x-1\right)^2=81\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=9\\x-1=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-8\end{matrix}\right.\)
a,\(\dfrac{1}{3}\times\left(x-1\right)+\dfrac{2}{5}\times\left(x+1\right)=0\)
b,4x-\(\left(x+\dfrac{1}{2}\right)=2x-\left(\dfrac{1}{2}x-5\right)\)
c,\(\left(x+\dfrac{1}{2}\right)\times\left(x-\dfrac{3}{4}\right)=0\)
Các bn ơi giúp mk với chiều mk đi học rồi !!!!!!!!!!
a. \(\dfrac{1}{3}.\left(x-1\right)+\dfrac{2}{5}.\left(x+1\right)=0\)
=> \(\dfrac{1}{3}x-\dfrac{1}{3}+\dfrac{2}{5}x+\dfrac{2}{5}=0\)
=> \(\dfrac{1}{3}x+\dfrac{2}{5}x=0+\dfrac{1}{3}-\dfrac{2}{5}\)
=> \(\dfrac{11}{15}x=\dfrac{-1}{15}\)
=> \(x=\dfrac{-1}{11}\)
Đây toán 8 mà? :v
a,\(\dfrac{1}{5}x\left(x-1\right)+\dfrac{2}{5}x\left(x+1\right)=0\)
\(\Leftrightarrow5x\left(x-1\right)+6x\left(x+1\right)=0\)
\(\Leftrightarrow\left[5\left(x-1\right)+6x\left(x+1\right)\right]x=0\)
\(\Leftrightarrow\left(5x-5+6x+6\right)x=0\)
\(\Leftrightarrow\left(11+1\right)x=0\)
\(\Leftrightarrow11x+1=0;x=0\)
\(\Leftrightarrow x=-\dfrac{1}{11};x=0\)
Vậy....
Bài 1 Tìm \(x\)
a)\(\left(\dfrac{1}{2}\times x-3\right)\times\left(-\dfrac{1}{3}+x\right)=0\)
b)\(\dfrac{1}{2}\times x^2-\dfrac{1}{5}\times x=0\)
c)\(\dfrac{1`}{4}\times x=\dfrac{1}{16}\times x^2\)
d)\(9\times x^2=1\)
e)\(\left(x-5\right)^2=4\)
a) \(\left(\dfrac{1}{2}x-3\right)\left(-\dfrac{1}{3}+x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-3=0\\-\dfrac{1}{3}+x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=0+3\\-\dfrac{1}{3}+x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3:\dfrac{1}{2}\\x=0-\left(-\dfrac{1}{3}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\dfrac{1}{3}\end{matrix}\right.\)
d) \(9x^2=1\)
\(\Leftrightarrow x^2=1:9\)
\(\Leftrightarrow x^2=\dfrac{1}{9}\)
\(\Leftrightarrow x^2=\left(\dfrac{1}{3}\right)^2\)
\(\Leftrightarrow x=\dfrac{1}{3}\)
e) \(\left(x-5\right)^2=4\)
\(\Leftrightarrow\left(x-5\right)^2=2^2\)
\(\Leftrightarrow x-5=2\)
\(\Leftrightarrow x=2+5\)
\(\Leftrightarrow x=7\)
Tìm x biết
a)\(\dfrac{2}{\left(x+2\right)\times\left(x+4\right)}+\dfrac{4}{\left(x+4\right)\times\left(x+8\right)}+\dfrac{6}{\left(x+8\right)\times\left(x+14\right)}=\dfrac{x}{\left(x+2\right)\times\left(x+14\right)}\)
Lời giải:
PT \(\Leftrightarrow \frac{(x+4)-(x+2)}{(x+2)(x+4)}+\frac{(x+8)-(x+4)}{(x+4)(x+8)}+\frac{(x+14)-(x+8)}{(x+8)(x+14)}=\frac{x}{(x+2)(x+14)}\)
\(\Leftrightarrow \frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+14}=\frac{x}{(x+2)(x+14)}\)
\(\Leftrightarrow \frac{1}{x+2}-\frac{1}{x+14}=\frac{x}{(x+2)(x+14)}\)
\(\Leftrightarrow \frac{12}{(x+2)(x+14)}=\frac{x}{(x+2)(x+14)}\)
\(\Rightarrow x=12\) (thỏa mãn)
Vậy......
>; <; =?
a) \(\dfrac{2}{3}\times\dfrac{4}{5}\) \(\dfrac{4}{5}\times\dfrac{2}{3}\)
b) \(\left(\dfrac{1}{3}\times\dfrac{2}{5}\right)\times\dfrac{3}{4}\) \(\dfrac{1}{3}\times\left(\dfrac{2}{5}\times\dfrac{3}{4}\right)\)
c) \(\left(\dfrac{1}{3}+\dfrac{2}{15}\right)\times\dfrac{3}{4}\) \(\dfrac{1}{3}\times\dfrac{3}{4}+\dfrac{2}{15}\times\dfrac{3}{4}\)
a) \(\dfrac{2}{3}\times\dfrac{4}{5}=\dfrac{4}{5}\times\dfrac{2}{3}\)
b) \(\left(\dfrac{1}{3}\times\dfrac{2}{5}\right)\times\dfrac{3}{4}=\dfrac{1}{3}\times\left(\dfrac{2}{5}\times\dfrac{3}{4}\right)\)
c) \(\left(\dfrac{1}{3}-\dfrac{2}{15}\right)\times\dfrac{3}{4}=\dfrac{1}{3}\times\dfrac{3}{4}+\dfrac{2}{15}\times\dfrac{3}{4}\)
thực hiện phép tính
a)\(\dfrac{2x^2-20x+50}{3x+3}\times\dfrac{x^2-1}{4\left(x-5\right)^2}\)
b) \(\dfrac{6x-3}{5x^2+x}\times\dfrac{25x^2+10x+1}{1-8x^3}\)
c) \(\dfrac{3x^2-x}{x^2-1}\times\dfrac{1-x^4}{\left(1-3x\right)^3}\)
a/ \(\dfrac{2x^2-20x+50}{3x+3}\cdot\dfrac{x^2-1}{4\left(x-5\right)^2}=\dfrac{2\left(x^2-10x+25\right)\cdot\left(x^2-1\right)}{3\left(x+1\right)\cdot4\left(x-5\right)^2}=\dfrac{2\left(x-5\right)^2\left(x-1\right)\left(x+1\right)}{12\left(x+1\right)\left(x-5\right)^2}=\dfrac{x+1}{6}\)
b/ \(\dfrac{6x-3}{5x^2+x}\cdot\dfrac{25x^2+10x+1}{1-8x^2}=-\dfrac{3\left(1-2x\right)\cdot\left(5x+1\right)^2}{x\left(5x+1\right)\left(1-2x\right)\left(1+2x+4x^2\right)}=\dfrac{3\left(5x+1\right)}{x\left(4x^2+2x+1\right)}\)
c/ \(\dfrac{3x^2-x}{x^2-1}\cdot\dfrac{1-x^4}{\left(1-3x\right)^3}=\dfrac{x-3x^2}{1-x^2}\cdot\dfrac{\left(1-x^2\right)\left(1+x^2\right)}{\left(1-3x\right)^3}=\dfrac{x\left(1-3x\right)\left(1-x^2\right)\left(1+x^2\right)}{\left(1-x^2\right)\left(1-3x\right)^3}=\dfrac{x\left(x^2+1\right)}{\left(1-3x\right)^3}\)
Dễ thế mà bạn ( người ko quen) ko làm đc !
Tìm GTLN của \(A=\dfrac{\left(x+2\right)^2}{2}\times\left(1-\dfrac{x^2}{x+2}\right)-\dfrac{x^2+6x+4}{x}\)
Bạn coi lại xem có viết nhầm chỗ nào trong biểu thức không? Biểu thức này nội việc rút gọn thôi đã "xấu" rồi.
\(B=\dfrac{\sqrt{x}\left(1-x\right)^2}{1+\sqrt{x}}\div\left[\left(\dfrac{x\sqrt{x}-1}{\sqrt{x}-1}+\sqrt{x}\right)\times\left(\dfrac{x\sqrt{x}+1}{\sqrt{x}+1}-\sqrt{x}\right)\right]\)
a) Rút gọn B
b) Tính B biết \(x=4-2\sqrt{3}\)
c) Tìm x để B = 2/3
d) So sánh B với 1
e) Tìm x để B>1
f) Tìm \(x\in Z\) để \(B\in Z\)
g) Tìm GTNN của B
h) Tìm m để B.m = 2 có nghiệm
ĐKXĐ: x>=0; x<>1
a: \(B=\dfrac{\sqrt{x}\left(x-1\right)^2}{\sqrt{x}+1}:\left(\left(x+\sqrt{x}+1+\sqrt{x}\right)\left(x-\sqrt{x}+1-\sqrt{x}\right)\right)\)
\(=\dfrac{\sqrt{x}\left(x-1\right)^2}{\sqrt{x}+1}:\left[\left(\sqrt{x}-1\right)^2\cdot\left(\sqrt{x}+1\right)^2\right]\)
\(=\dfrac{\sqrt{x}\left(x-1\right)^2}{\left(x-1\right)^2\cdot\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}}{\sqrt{x}+1}\)
b: Khi x=4-2căn 3=(căn 3-1)^2 thì \(B=\dfrac{\sqrt{3}-1}{\sqrt{3}-1+1}=\dfrac{\sqrt{3}-1}{\sqrt{3}}=\dfrac{3-\sqrt{3}}{3}\)
c: B=2/3
=>căn x/căn x+1=2/3
=>căn x=2
=>x=4
d: \(B-1=\dfrac{\sqrt{x}-\sqrt{x}-1}{\sqrt{x}+1}=-\dfrac{1}{\sqrt{x}+1}< 0\)
=>B<1
e: B>1
=>-1/căn x+1>0
=>căn x+1<0(vô lý)
=>KO có x thỏa mãn
f: B nguyên khi căn x chia hết cho căn x+1
=>căn x+1-1 chia hết cho căn x+1
=>căn x+1=1 hoặc căn x+1=-1(loại)
=>căn x=0
=>x=0