Tìm x:
a)(2x- 6)7 = (2x-6)5
b) (7- 2x)9 = (7- 2x)3
K
Khách
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9 tháng 1 2022
a: \(\Leftrightarrow-2x=-\dfrac{1}{3}-\dfrac{1}{8}-\dfrac{5}{7}=-\dfrac{197}{168}\)
hay x=197/336
c: \(\Leftrightarrow5x=9+\dfrac{6}{18}-\dfrac{2}{7}=\dfrac{190}{21}\)
hay x=38/21
NM
2
28 tháng 8 2021
a: Ta có: \(4\left(2-x\right)+x\left(x+6\right)=x^2\)
\(\Leftrightarrow8-4x+x^2+6x-x^2=0\)
\(\Leftrightarrow2x=-8\)
hay x=-4
b: Ta có: \(x\left(x-7\right)-\left(x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow x^2-7x-x^2-3x+10=0\)
\(\Leftrightarrow-10x=-10\)
hay x=1
c: Ta có: \(\left(2x+3\right)\left(3-2x\right)+\left(2x-1\right)^2=2\)
\(\Leftrightarrow9-4x^2+4x^2-4x+1=2\)
\(\Leftrightarrow-4x=-8\)
hay x=2
L
19 tháng 5 2022
a) (2x - 3)(6 - 2x) = 0
=> \(\left[{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.=>\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=3\end{matrix}\right.\)
b) \(5\dfrac{4}{7}:x=13=>\dfrac{39}{7}:x=13=>x=\dfrac{39}{7}:13=>x=\dfrac{3}{7}\)
c) \(2x-\dfrac{3}{7}=6\dfrac{2}{7}=>2x-\dfrac{3}{7}=\dfrac{44}{7}=>2x=\dfrac{47}{7}=>x=\dfrac{47}{14}\)
d) \(\dfrac{x}{5}+\dfrac{1}{2}=\dfrac{6}{10}=>\dfrac{x}{5}=\dfrac{6}{10}-\dfrac{1}{2}=>\dfrac{x}{5}=\dfrac{1}{10}=>x.10=5=>x=\dfrac{1}{2}\)
e) \(\dfrac{x+3}{15}=\dfrac{1}{3}=>\left(x+3\right).3=15=>x+3=5=>x=2\)
27 tháng 7 2021
a) Ta có: \(\left(2x-5\right)^3=216\)
\(\Leftrightarrow2x-5=6\)
\(\Leftrightarrow2x=11\)
hay \(x=\dfrac{11}{2}\)
b) Ta có: \(2x-3⋮x+4\)
\(\Leftrightarrow-11⋮x+4\)
\(\Leftrightarrow x+4\in\left\{1;-1;11;-11\right\}\)
hay \(x\in\left\{-3;-5;7;-15\right\}\)
3 tháng 12 2023
Alo, sugeni two wai phem. Si ga no, you woo be the me that nas te, ai gi da
C
1
7 tháng 7 2021
\(a,\left(3x-7\right)^2=\left(2-2x\right)^2\)
a,\(=>\left(3x-7\right)^2-\left(2-2x\right)^2=0\)
\(< =>\left(3x-7+2-2x\right)\left(3x-7-2+2x\right)=0\)
\(< =>\left(x-5\right)\left(5x-9\right)=0=>\left[{}\begin{matrix}x=5\\x=1,8\end{matrix}\right.\)
b, \(x^2-8x+6=0< =>x^2-2.4x+16-10=0\)
\(< =>\left(x-4\right)^2-\sqrt{10}^2=0\)
\(=>\left(x-4+\sqrt{10}\right)\left(x-4-\sqrt{10}\right)=0\)
\(=>\left[{}\begin{matrix}x=4-\sqrt{10}\\x=4+\sqrt{10}\end{matrix}\right.\)
c, \(4x^2-2x-1=0\)
\(< =>\left(2x\right)^2-2.2.\dfrac{1}{2}x+\dfrac{1}{4}-\dfrac{5}{4}=0\)
\(=>\left(2x-\dfrac{1}{2}\right)^2-\left(\dfrac{\sqrt{5}}{2}\right)^2=0\)
\(=>\left(2x+\dfrac{-1+\sqrt{5}}{2}\right)\left(2x-\dfrac{1+\sqrt{5}}{2}\right)=0\)
\(=>\left[{}\begin{matrix}x=\dfrac{1-\sqrt{5}}{4}\\x=\dfrac{1+\sqrt{5}}{4}\end{matrix}\right.\)
d,\(x^4-4x^2-32=0\)
đặt \(t=x^2\left(t\ge0\right)=>t^2-4t-32=0\)
\(< =>t^2-2.2t+4-6^2=0\)
\(=>\left(t-2\right)^2-6^2=0=>\left(t-8\right)\left(t+4\right)=0\)
\(=>\left[{}\begin{matrix}t=8\left(tm\right)\\t=-4\left(loai\right)\end{matrix}\right.\)\(=>x=\pm\sqrt{8}\)
VN
Võ Ngọc Phương
VIP
11 tháng 9 2023
Bài 4:
a) \(\dfrac{4}{3}+\left(1,25-x\right)=2,25\)
\(1,25-x=2,25-\dfrac{4}{3}=\dfrac{9}{4}-\dfrac{4}{3}\)
\(1,25-x=\dfrac{11}{12}\)
\(x=1,25-\dfrac{11}{12}=\dfrac{5}{4}-\dfrac{11}{12}\)
\(x=\dfrac{1}{3}\)
b) \(\dfrac{17}{6}-\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)
\(x-\dfrac{7}{6}=\dfrac{17}{6}-\dfrac{7}{4}=\dfrac{34}{12}-\dfrac{21}{12}\)
\(x-\dfrac{7}{6}=\dfrac{13}{12}\)
\(x=\dfrac{13}{12}+\dfrac{7}{6}=\dfrac{13}{12}+\dfrac{14}{12}\)
\(x=\dfrac{27}{12}=\dfrac{9}{4}\)
c) \(4-\left(2x+1\right)=3-\dfrac{1}{3}=\dfrac{9}{3}-\dfrac{1}{3}\)
\(4-\left(2x+1\right)=\dfrac{8}{3}\)
\(2x+1=\dfrac{8}{3}+4=\dfrac{8}{3}+\dfrac{12}{3}\)
\(2x+1=\dfrac{20}{3}\)
\(2x=\dfrac{20}{3}-1=\dfrac{20}{3}-\dfrac{3}{3}\)
\(2x=\dfrac{17}{3}\)
\(x=\dfrac{17}{3}.\dfrac{1}{2}=\dfrac{17}{6}\)
Bài 15:
a) \(\left(\dfrac{-2}{3}\right)^9:x=\dfrac{-2}{3}\)
\(x=\left(\dfrac{-2}{3}\right)^9:\dfrac{-2}{3}=\left(\dfrac{-2}{3}\right)^{9-1}\)
\(=>x=\left(\dfrac{-2}{3}\right)^8\)
b) \(x:\left(\dfrac{4}{9}\right)^5=\left(\dfrac{4}{9}\right)^4\)
\(x=\left(\dfrac{4}{9}\right)^4.\left(\dfrac{4}{9}\right)^5=\left(\dfrac{4}{9}\right)^{4+5}\)
\(=>x=\left(\dfrac{4}{9}\right)^9\)
c) \(\left(x+4\right)^3=-125\)
\(\left(x+4\right)^3=\left(-5\right)^3\)
\(=>x+4=-5\)
\(x=-5-4\)
\(=>x=-9\)
d) \(\left(10-5x\right)^3=64\)
\(\left(10-5x\right)^3=4^3\)
\(=>10-5x=4\)
\(5x=10-4\)
\(5x=6\)
\(=>x=\dfrac{6}{5}\)
e) \(\left(4x+5\right)^2=81\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(4x+5\right)^2=\left(-9\right)^2\\\left(4x+5\right)^2=9^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+5=-9\\4x+5=9\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=-14\\4x=4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-14}{4}\\x=1\end{matrix}\right.\)
Bài 16:
a) \(4-1\dfrac{2}{5}-\dfrac{8}{3}\)
\(=4-\dfrac{7}{5}-\dfrac{8}{3}\)
\(=\dfrac{60-21-40}{15}=\dfrac{-1}{15}\)
b) \(-0,6-\dfrac{-4}{9}-\dfrac{16}{15}\)
\(=\dfrac{-3}{5}+\dfrac{4}{9}-\dfrac{16}{15}\)
\(=\dfrac{\left(-27\right)+20-48}{45}=\dfrac{-55}{45}=\dfrac{-11}{9}\)
c) \(-\dfrac{15}{4}.\left(\dfrac{-7}{15}\right).\left(-2\dfrac{2}{5}\right)\)
\(=\dfrac{7}{4}.\dfrac{-12}{5}\)
\(=\dfrac{-21}{5}\)
\(#Wendy.Dang\)
DT
Dang Tung
CTVHS
31 tháng 1 2023
\(< =>\dfrac{13\left(x+3\right)}{\left(2x+7\right)\left(x-3\right)\left(x+3\right)}+\dfrac{x^2-9}{\left(2x+7\right)\left(x-3\right)\left(x+3\right)}=\dfrac{6\left(2x+7\right)}{\left(2x+7\right)\left(x-3\right)\left(x+3\right)}\left(ĐK:x\ne\left\{-\dfrac{7}{2};3;-3\right\}\right)\\ =>13x+39+x^2-9=12x+42\\ < =>x^2+x-12=0\\ < =>\left(x+4\right)\left(x-3\right)=0\\ =>\left[{}\begin{matrix}x=-4\left(TM\right)\\x=3\left(KTM\right)\end{matrix}\right.\\ =>S=\left\{-4\right\}\)
31 tháng 1 2023
\(ĐKXĐ:x\ne\dfrac{7}{2}\) và \(x\ne\pm3\)
mẫu chung : \(\left(2x+7\right)\left(x+3\right)\left(x-3\right)\)
Khử mẫu ta được :
\(13\left(x+3\right)+\left(x+3\right)\left(x-3\right)=6\left(2x+7\right)\)
\(\Leftrightarrow x^2+x-12=0\)
\(\Leftrightarrow\left(x+4\right)\left(x-3\right)=0\)
\(x=\left\{{}\begin{matrix}-4\\3\end{matrix}\right.\)
do \(x=3\) không thỏa mãn điều kiện thích hợp nên pt có nghiệm duy nhất là : \(-4\)
\(Vậy...\)
LN
15 tháng 4 2020
1) (x+6)(3x-1)+x+6=0
⇔(x+6)(3x-1)+(x+6)=0
⇔(x+6)(3x-1+1)=0
⇔3x(x+6)=0
2) (x+4)(5x+9)-x-4=0
⇔(x+4)(5x+9)-(x+4)=0
⇔(x+4)(5x+9-1)=0
⇔(x+4)(5x+8)=0
3)(1-x)(5x+3)÷(3x-7)(x-1)
=\(\frac{\left(1-x\right)\left(5x+3\right)}{\left(3x-7\right)\left(x-1\right)}=\frac{\left(1-x\right)\left(5x+3\right)}{\left(7-3x\right)\left(1-x\right)}=\frac{\left(5x+3\right)}{\left(7-3x\right)}\)
NT
24 tháng 8 2021
a.ĐKXĐ: \(x\ge0\)
\(\sqrt{2x}< \dfrac{1}{3}\) \(\Leftrightarrow2x< \dfrac{1}{3}\Leftrightarrow6x< 1\Leftrightarrow x< \dfrac{1}{6}\)
b. ĐKXĐ: \(x\ge\dfrac{1}{6}\)
\(\sqrt{-3x+\dfrac{1}{2}}\ge5\Leftrightarrow-3x+\dfrac{1}{2}\ge25\Leftrightarrow x=-\dfrac{49}{6}\)
c. ĐKXĐ: \(x\ge\dfrac{1}{2}\)
\(\sqrt{-2x+1}>7\) \(\Leftrightarrow-2x+1>49\Leftrightarrow x=-24\)
d. ĐKXĐ: \(x\ge\dfrac{1}{2}\)
\(\sqrt{2x-1}\le\dfrac{3}{2}\Leftrightarrow2x-1\le\dfrac{9}{4}\Leftrightarrow x=\dfrac{13}{8}\)
25 tháng 8 2021
a: Ta có: \(\sqrt{2x}< \dfrac{1}{3}\)
\(\Leftrightarrow2x< \dfrac{1}{9}\)
\(\Leftrightarrow x< \dfrac{1}{18}\)
Kết hợp ĐKXĐ, ta được: \(0\le x< \dfrac{1}{18}\)
b: Ta có: \(\sqrt{-3x+\dfrac{1}{2}}\ge5\)
\(\Leftrightarrow-3x+\dfrac{1}{2}\ge25\)
\(\Leftrightarrow-3x\ge\dfrac{49}{2}\)
hay \(x\le-\dfrac{49}{6}\)
c: Ta có: \(\sqrt{-2x+1}>7\)
\(\Leftrightarrow-2x+1>49\)
\(\Leftrightarrow-2x>48\)
hay x<-24
a)vi (2x-6)^7=(2x-6)^5
=>2x-6=0 hoac 2x-6=1
=>x=3hoac x=7/2
b)(7-2x)^9=(7-2x)^3
=>7-2x=0 hoac 7-2x=1
=>x=7/2 hoac x=3
a) \(\left(2x-6\right)^7=\left(2x-6\right)^5\)
\(\Leftrightarrow\left[2\left(x-3\right)\right]^7=\left[2\left(x-3\right)\right]^5\)
\(\Leftrightarrow2^7\left(x-3\right)^7=2^5\left(x-3\right)^5\)
\(\Leftrightarrow128\left(x-3\right)^7=32\left(x-3\right)^5\)
\(\Leftrightarrow4\left(x-3\right)^7=\left(x-3\right)^5\)
\(\Leftrightarrow4\left(x-3\right)^7-\left(x-3\right)^5=0\)
\(\Leftrightarrow\left(x-3\right)^5\left[4\left(x-3\right)-1\right]=0\)
\(\Leftrightarrow\hept{\begin{cases}x-3=0\\4\left(x-3\right)-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=3\\x=\frac{7}{2}\\x=\frac{5}{2}\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=3\\x=\frac{7}{2}\\x=\frac{5}{2}\end{cases}}\)