Tìm x, biết:
a. (7x - 11)3 = 25 . 52 + 200
B. 3 với 1/3x + 16 với 3/4 = -13,25
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13 tháng 8 2021
a: Ta có: \(7x+25=144\)
\(\Leftrightarrow7x=119\)
hay x=17
b: Ta có: \(33-12x=9\)
\(\Leftrightarrow12x=24\)
hay x=2
c: Ta có: \(128-3\left(x+4\right)=23\)
\(\Leftrightarrow3\left(x+4\right)=105\)
\(\Leftrightarrow x+4=35\)
hay x=31
d: Ta có: \(71+\left(726-3x\right)\cdot5=2246\)
\(\Leftrightarrow5\left(726-3x\right)=2175\)
\(\Leftrightarrow726-3x=435\)
\(\Leftrightarrow3x=291\)
hay x=97
e: Ta có: \(720:\left[41-\left(2x+5\right)\right]=40\)
\(\Leftrightarrow41-\left(2x+5\right)=18\)
\(\Leftrightarrow2x+5=23\)
\(\Leftrightarrow2x=18\)
hay x=9
18 tháng 3 2018
a) (7x - 11)3 = 25 x 52 + 200
(7x - 11)3 = 800 + 200
(7x - 11)3 = 1000
(7x - 11)3 = 103
=> 7x - 11 = 10
=> 7x = 10 + 11
=> 7x = 21
=> x = 3
b) \(3\frac{1}{3}x+16\frac{3}{4}=-13,25\)
\(3\frac{1}{3}x=-13,25-16\frac{3}{4}\)
\(\frac{10}{3}x=-30\)
\(x=-9\)
TN
3
12 tháng 2 2017
\(a,3\frac{1}{3}x+16\frac{3}{4}=-13,25\)
\(\frac{10}{3}x+\frac{67}{4}=-13,25\)
\(\frac{10}{3}x=-13,25-\frac{67}{4}\)
\(\frac{10}{3}x=-30\)
\(x=\left(-30\right):\frac{10}{3}\)
\(x=-9\)
\(b,\left(7x-11\right)^3=2^5.5^2+200\)
\(\left(7x-11\right)^3=32.25+200\)
\(\left(7x-11\right)^3=1000\)
\(\Rightarrow\left(7x-11\right)^3=10^3\)
\(\Rightarrow7x-11=10\)
\(\Rightarrow7x=10+11\)
\(\Rightarrow7x=21\)
\(\Rightarrow x=3\)
12 tháng 2 2017
A) (10/3)x+67/4=-53/4<=>(10/3)x=-53/4-67/4=-30<=>x=-30:(10/3)=-9 b) (7x-11)^3=1000=10^3<=>7x-11=10=>7x=21=>x=3
11 tháng 7 2021
9) Ta có: \(\dfrac{2x+5}{x+3}+1=\dfrac{4}{x^2+2x-3}-\dfrac{3x-1}{1-x}\)
\(\Leftrightarrow\left(2x+5\right)\left(x-1\right)+x^2+2x-3=4+\left(3x-1\right)\left(x+3\right)\)
\(\Leftrightarrow2x^2-2x+5x-5+x^2+2x-3-4-3x^2-10x+x+3=0\)
\(\Leftrightarrow-4x=9\)
hay \(x=-\dfrac{9}{4}\)
11 tháng 7 2021
10) Ta có: \(\dfrac{x-1}{x+3}-\dfrac{x}{x-3}=\dfrac{7x-3}{9-x^2}\)
\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3-7x}{\left(x-3\right)\left(x+3\right)}\)
Suy ra: \(x^2-4x+3-x^2-3x-3+7x=0\)
\(\Leftrightarrow0x=0\)(luôn đúng)
Vậy: S={x|\(x\notin\left\{3;-3\right\}\)}
11) Ta có: \(\dfrac{5+9x}{x^2-16}=\dfrac{2x-1}{x+4}+\dfrac{3x-1}{x-4}\)
\(\Leftrightarrow\dfrac{\left(2x-1\right)\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}+\dfrac{\left(3x-1\right)\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{9x+5}{\left(x-4\right)\left(x+5\right)}\)
Suy ra: \(2x^2-9x+4+3x^2+12x-x-4-9x-5=0\)
\(\Leftrightarrow5x^2-7x=0\)
\(\Leftrightarrow x\left(5x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{7}{5}\end{matrix}\right.\)
12) Ta có: \(\dfrac{2x}{2x-1}+\dfrac{x}{2x+1}=1+\dfrac{4}{\left(2x-1\right)\left(2x+1\right)}\)
\(\Leftrightarrow\dfrac{2x\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)}+\dfrac{x\left(2x-1\right)}{\left(2x+1\right)\left(2x-1\right)}=\dfrac{4x^2-1+4}{\left(2x-1\right)\left(2x+1\right)}\)
Suy ra: \(4x^2+2x+2x^2-x-4x^2-3=0\)
\(\Leftrightarrow2x^2+x-3=0\)
\(\Leftrightarrow2x^2+3x-2x-3=0\)
\(\Leftrightarrow\left(2x+3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=1\end{matrix}\right.\)
a) (7x - 11)3 = 25.52 + 200
(7x - 11)3 = 32.25 + 200
(7x - 11)3 = 800 + 200
(7x - 11)3 = 1000 = 103
=> 7x - 11 = 10
=> 7x = 10 + 11 = 21
=> x = 21 : 7 = 3
b) \(1\frac{1}{3}x+16\frac{3}{4}=-13,25\)
=> \(\frac{4}{3}x+\frac{67}{4}=-\frac{53}{4}\)
=> \(\frac{4}{3}x=-\frac{53}{4}-\frac{67}{4}\)
=> \(\frac{4}{3}x=-30\)
=> \(x=-30:\frac{4}{3}\)
=> \(x=-30.\frac{3}{4}=-\frac{45}{2}\)