Tìm x
\(\left(x^4\right)^2=\frac{x^{12}}{x^5}\left(x\ne0\right)\)
\(\left(3x-1\right)^3=\frac{8}{27}\)
\(a.\frac{x-1}{x+2}=\frac{4}{5}\left(x\ne-2\right)\) e)\(\frac{x}{-8}=\frac{2}{-x^3}\left(x\ne0\right)\)
b)\(\frac{1}{12}:\frac{4}{21}=3\frac{1}{2}:\left(3x-2\right)\) F)\(\frac{x}{8}=\frac{x}{x^3}\left(x\ne0\right)\)
c)\(\frac{x^2}{-8}=\frac{27}{x}\left(x\ne0\right)\)
d)\(\frac{x+7}{-20}=\frac{-5}{x+7}\left(x\ne-7\right)\)
\(a.\frac{x-1}{x+2}=\frac{4}{5}\)
\(\Rightarrow\frac{x+2-3}{x+2}=\frac{4}{5}\)
\(\Rightarrow1-\frac{3}{x+2}=\frac{4}{5}\)
\(\Rightarrow\frac{3}{x+2}=1-\frac{4}{5}\)
\(\Rightarrow\frac{3}{x+2}=\frac{1}{5}\)
\(\Rightarrow\frac{3}{x+2}=\frac{3}{15}\Rightarrow x+2=15\)
\(\Rightarrow x=13\)( thỏa mãn )
tìm x biết
a) \(\frac{x-1}{x+2}=\frac{4}{5}\left(x\ne-2\right)\) b)22x+1+4x+3=264 c)\(\frac{x^2}{-8}=\frac{27}{x}\left(x\ne0\right)\) d)\(\frac{x+7}{-20}=\frac{-5}{x+7}\left(x\ne-7\right)\) e)\(\frac{x}{-8}=\frac{2}{-x^3}\left(x\ne0\right)\)
a)Ta có:
\(\frac{x-1}{x+2}=\frac{4}{5}\Leftrightarrow5\left(x-1\right)=4\left(x+2\right)\)
\(\Leftrightarrow5x-5=4x+8\)
\(\Leftrightarrow5x-4x=8+5\)
\(\Leftrightarrow x=13\)
b)Ta có:
\(2^{2x+1}+4^{x+3}=2^{2x+1}+2^{2x+6}=2^{2x+1}\left(1+2^5\right)=2^{2x+1}.33=264\Leftrightarrow2^{2x+1}=8=2^3\)\(\Rightarrow2x+1=3\Leftrightarrow2x=2\Leftrightarrow x=1\)
c)Ta có:
\(\frac{x^2}{-8}=\frac{27}{x}\Leftrightarrow x^3=-8.27=-216\Leftrightarrow x=-6\)
d)Ta có:
\(\frac{x+7}{-20}=\frac{-5}{x+7}\Leftrightarrow\left(x+7\right)^2=\left(-20\right)\left(-5\right)=100\Leftrightarrow\left[{}\begin{matrix}x+7=10\\x+7=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-17\end{matrix}\right.\)e)Ta có:
\(\frac{x}{-8}=\frac{2}{-x^3}\Leftrightarrow x.\left(-x^3\right)=-8.2\)
\(\Leftrightarrow-x^4=-16\Leftrightarrow x^4=16\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
1 Tìm x:
( \(3x-2\frac{1}{3}\)):( \(3\frac{1}{4}-5\frac{2}{3}+1\frac{4}{5}\)) = \(2-1\frac{1}{3}x\)
2. Tìm x:
\(\frac{x+\frac{2\left(3-x\right)}{5}}{14}-\frac{5x-4\left(x-1\right)}{24}=\frac{7x+2+\frac{9-3x}{5}}{12}+\frac{2}{3}\)
3. Tìm x:
\(\left(1+3x\right)^2-3x\left(2x+6\right)=\left(4-3x\right)\left(x+3\right)-\left(2x-1\right)^2\)
1) \(x=\frac{99}{196}\)
2) \(x=-2\)
3) \(x\approx-0,59\)
1)2x(25x-4)-(5x-2)(5x+1)=8 / 5)\(2\left(x-2\right)-3\left(3x-1\right)=\left(x-3\right)\)
2)x(4x-3)-(2x-2)(2x-1)=5 / 6)\(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{\left(x+1\right)\left(x-2\right)}\)
3)\(\frac{5}{2x+3}+\frac{3}{9-x^2}=\frac{8}{7\left(x=3\right)}\) / 7)\(\frac{5x-2}{6}+\frac{3-4x}{2}=2-\frac{x+7}{3}\)
4)\(\frac{2}{3\left(x-2\right)}+\frac{5}{12-3x^2}=\frac{3}{4\left(x+2\right)}\) / 8)\(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{\left(x+1\right)\left(x-2\right)}\)
Đây là lớp 8 nha các b giúp mk với
Do mk viết nhầm
Tìm x \(\in\) Z
a,\(\frac{1}{2}-\left(\frac{1}{3}+\frac{1}{4}\right)< x< \frac{1}{48}-\left(\frac{1}{16}-\frac{1}{6}\right)\)
b, \(\frac{3}{4}-\frac{5}{6}\le\frac{x}{12}< 1-\left(\frac{2}{3}-\frac{1}{3}\right)\)
c, \(x:\left(\frac{1}{2}\right)^2=\frac{-1}{2}\)
d,\(\left(x^4\right)^2=\frac{x^{12}}{x^5}\left(x\ne0\right)\)
e,\(\frac{3^2.3^8}{27^3}=3^x\)
f,\(2^{x-1}=\left(16\right)^5\)
a: \(\Leftrightarrow\dfrac{1}{2}-\dfrac{7}{12}< x< \dfrac{1}{48}+\dfrac{5}{48}=\dfrac{6}{48}=\dfrac{1}{8}\)
\(\Leftrightarrow-\dfrac{1}{12}< x< \dfrac{1}{8}\)
=>x=0
c: \(\Leftrightarrow x=\dfrac{-1}{2}\cdot\dfrac{1}{4}=\dfrac{-1}{8}\)
d: \(\Leftrightarrow x^8=x^7\)
=>x(x-1)=0
=>x=0(loại) hoặc x=1(nhận)
e: \(\Leftrightarrow3^x=\dfrac{3^{10}}{3^9}=3\)
hay x=1
f: =>x-1=20
hay x=21
Tìm x : \(\frac{2\left(x-1\right)\left(x-3\right)}{3}-\frac{4\left(2x-1\right)^2}{5}=\frac{\left(1+3x\right)^2}{2}-3x\left(1-x\right)\)
\(\Leftrightarrow20\left(x^2-4x+3\right)-24\left(4x^2-4x+1\right)=15\left(9x^2+6x+1\right)+90x\left(x-1\right)\)
\(\Leftrightarrow20x^2-80x+60-96x^2+96x-24=135x^2+90x+15+90x^2-90x\)
\(\Leftrightarrow-301x^2+16x+21=0\)
\(\text{Δ}=16^2-4\cdot\left(-301\right)\cdot21=25540\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là
\(\left\{{}\begin{matrix}x_1=\dfrac{-16-\sqrt{25540}}{-602}=\dfrac{16+\sqrt{25540}}{602}\\x_2=\dfrac{16-\sqrt{25540}}{602}\end{matrix}\right.\)
Tìm x : \(\frac{2\left(x-1\right)\left(x-3\right)}{3}-\frac{4\left(2x-1\right)^2}{5}=\frac{\left(1+3x\right)^2}{2}-3x\left(1-x\right)\)
Tìm x: \(\frac{\left(x-2\right)^2}{2}-4\frac{1}{3}\left(x+3\right)^2=\frac{1}{4}\left(x-1\right)\left(x-2\right)-2\left(3x-1\right)\left(2+3x\right)\)
\(\Leftrightarrow\dfrac{1}{2}\left(x^2-4x+4\right)-\dfrac{13}{3}\left(x^2+6x+9\right)=\dfrac{1}{4}\left(x^2-3x+2\right)-2\left(9x^2+3x-2\right)\)
\(\Leftrightarrow x^2\cdot\dfrac{1}{2}-2x+2-\dfrac{13}{3}x^2-26x-39=\dfrac{1}{4}x^2-\dfrac{3}{4}x+\dfrac{1}{2}-18x^2-6x+4\)
\(\Leftrightarrow x^2\cdot\dfrac{167}{12}-\dfrac{85}{4}x-\dfrac{83}{2}=0\)
\(\Leftrightarrow167x^2-255x-498=0\)
\(\text{Δ}=\left(-255\right)^2-4\cdot167\cdot\left(-498\right)=397689\)
Vì Δ>0 nên phương trình có 2 nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{255-\sqrt{397689}}{334}\\x_2=\dfrac{255+\sqrt{397689}}{334}\end{matrix}\right.\)
1. \(\frac{7}{8}x-5\left(x-9\right)=\frac{20x+1,5}{6}\)
2 . \(\frac{\left(2x+1\right)^2}{5}-\frac{\left(x+1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)
3 . \(4\left(3x-2\right)-3\left(x-4\right)=7x+10\)
4. \(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}\)
1) \(\frac{7}{8}x-5\left(x-9\right)=\frac{20x+1,5}{6}\)
<=> \(\frac{21x}{24}-\frac{100\left(x-9\right)}{24}=\frac{80x+6}{24}\)
<=> 21x - 100x + 900 = 80x + 6
<=> -79x - 80x = 6 - 900
<=> -159x = -894
<=> x = 258/53
Vậy S = {258/53}
2) \(\frac{\left(2x+1\right)^2}{5}-\frac{\left(x+1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)
<=> \(\frac{3\left(4x^2+4x+1\right)}{15}-\frac{5\left(x^2+2x+1\right)}{15}=\frac{7x^2-14x-5}{15}\)
<=> 12x2 + 12x + 3 - 5x2 - 10x - 5 = 7x2 - 14x - 5
<=> 7x2 + 2x - 7x2 + 14x = -5 + 2
<=> 16x = 3
<=> x = 3/16
Vậy S = {3/16}
3) 4(3x - 2) - 3(x - 4) = 7x+ 10
<=> 12x - 8 - 3x + 12 = 7x + 10
<=> 9x - 7x = 10 - 4
<=> 2x = 6
<=> x = 3
Vậy S = {3}
4) \(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}\)
<=> \(\frac{x^2+14x+40}{12}+\frac{3\left(x^2+2x-8\right)}{12}=\frac{4\left(x^2+8x-20\right)}{12}\)
<=> x2 + 14x + 40 + 3x2 + 6x - 24 = 4x2 + 32x - 80
<=> 4x2 + 20x - 4x2 - 32x = -80 - 16
<=> -12x = -96
<=> x = 8
Vậy S = {8}