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1) /7x+14/=63
/7x/ =63-14
/x/ =49:7
x =7
2) /18+9x/=81
/9x/=81-18
/x/=63:9
x=7
a) -5 . (2 - x) + 4(x - 3) = 10x - 15
-10 + 5x + 4x -12 = 10x - 15
5x + 4x - 10x = -15 + 10 + 12
-x = 7
x = -7
b) 5 . (3 - 2x) + 5 . (x - 4) = 6 - 4x
15 - 10x + 5x - 20 = 6 - 4x
-10x + 5x + 4x = 6 - 15 + 20
-x = 11
x = -11
c) - 7 . (3x - 5) + 2 . (7x - 14) = 28
-21x + 35 + 14x - 28 = 28
-21x + 14x = 28 - 35 + 28
-7x = 21
x = 21 : (-7)
x = -3
d) 4 . (x - 5) - 3 . (x + 7) = 5 . (-4)
4x - 20 - 3x - 21 = -20
4x - 3x = -20 + 20 + 21
x = 21
e) 5 . (4 - x) - 7. (-x + 2) = 4 - 9 + 3
20 - 5x + 7x - 14 = -2
-5x + 7x = -2 - 20 + 14
2x = -8
x = -8 : 2
x = -4
Đúng 100%
câu c
- 7 ( 3x - 5 ) + 2 ( 7x - 14 ) = 28
- 21x + 35 + 14x - 28 = 28
21x - 14x = 35 - 28 - 28
7x = - 21
x = ( - 21) : 7
x = - 3
a, =>-2x+12+6x-60=8
=>4x+48=8
=>4x=56
=>x=14
b,=>-8x-36-8x-3-x-13=0
=>-17x-52=0
=>-17x=52
=>x=-52/17
c,=>14x+7x^2-7x^2-21x=14
=>-7x=14
=>x=-2
\(a.\dfrac{3}{2}+\dfrac{-1}{3}< \dfrac{x}{6}< \dfrac{1}{9}+\dfrac{31}{18}\)
\(\Leftrightarrow\dfrac{7}{6}< \dfrac{x}{6}< \dfrac{11}{6}\)
\(\Leftrightarrow7< x< 11\)
\(\Leftrightarrow x\in\left\{8;9;10\right\}\)
\(b.\dfrac{-5}{12}+\dfrac{7}{12}+\dfrac{-1}{12}< \dfrac{x}{12}< \dfrac{2}{15}+\dfrac{1}{5}\)
\(\Leftrightarrow\dfrac{1}{12}< \dfrac{x}{12}< \dfrac{1}{3}\)
\(\Leftrightarrow\dfrac{1}{12}< \dfrac{x}{12}< \dfrac{4}{12}\)
\(\Leftrightarrow1< x< 4\)
\(\Leftrightarrow x\in\left\{2;3\right\}\)
a) Ta có: \(\left|-5\right|+\left|x-1\right|=\left|7\right|\)
\(\Leftrightarrow\left|x-1\right|+5=7\)
\(\Leftrightarrow\left|x-1\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=2\\x-1=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
Vậy: \(x\in\left\{3;-1\right\}\)
b) Ta có: \(2\cdot\left|2x-4\right|-\left|-4\right|=\left|-50\right|\)
\(\Leftrightarrow4\cdot\left|x-2\right|-4=50\)
\(\Leftrightarrow4\cdot\left|x-2\right|=54\)
\(\Leftrightarrow\left|x-2\right|=\dfrac{27}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=\dfrac{27}{2}\\x-2=-\dfrac{27}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{31}{2}\left(loại\right)\\x=-\dfrac{23}{2}\left(loại\right)\end{matrix}\right.\)
Vậy: \(x\in\varnothing\)
a, | -5 | + | x-1 | = | 7 |
5 + | x - 1 | = 7
| x - 1 | = 2
TH1 x -1 = 2
x = 3
TH2 x -1 = -2
x= -1
\(\frac{6}{11}\cdot\frac{3}{14}+\frac{-5}{16}+\frac{3}{14}\cdot\frac{5}{11}+\frac{-3}{16}\)
\(=\frac{6}{11}\cdot\frac{3}{14}+\frac{3}{14}\cdot\frac{5}{11}+\frac{-5}{16}+\frac{-3}{16}\)
\(=\frac{3}{14}\cdot\left(\frac{6}{11}+\frac{5}{11}\right)+\left(\frac{-5}{16}+\frac{-3}{16}\right)\)
\(=\frac{3}{14}\cdot\frac{6+5}{11}+\frac{-5+\left(-3\right)}{16}\)
\(=\frac{3}{14}\cdot\frac{11}{11}+\frac{-8}{16}\)
\(=\frac{3}{14}\cdot1+\frac{-1}{2}\)
\(=\frac{3}{14}+\frac{-1}{2}\)
\(=\frac{3}{14}+\frac{-7}{14}\)
\(=\frac{3+\left(-7\right)}{14}\)\(=\frac{-4}{14}=\frac{-2}{7}\)
\(\frac{-5}{6}+\left(7x-\frac{1}{2}\right)\cdot\frac{2}{9}=-1\frac{1}{3}\)
\(\frac{-5}{6}+\frac{14}{9}\cdot x-\frac{1}{9}=-\frac{4}{3}\)
\(\frac{14}{9}\cdot x-\frac{1}{9}=-\frac{4}{3}+\frac{5}{6}\)
\(\frac{14}{9}\cdot x-\frac{1}{9}=-\frac{1}{2}\)
\(\frac{14}{9}\cdot x=-\frac{1}{2}+\frac{1}{9}\)
\(\frac{14}{9}\cdot x=-\frac{7}{18}\)
\(x=-\frac{7}{18}:\frac{14}{9}\)
\(x=-\frac{1}{4}\)
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Các bn ơi giúp mình câu này đc ko mong các bạn giúp đỡ 4(x-1)=2x-6(x-2) các bạn giải hộ mình với
Bài 2
\(a,\left(x-5\right)^4=\left(x-5\right)^6\)
\(\Rightarrow\left(x-5\right)^6-\left(x-5\right)^4=0\)
\(\Rightarrow\left(x-5\right)^4\left[\left(x-5\right)^2-1\right]=0\)
\(\Rightarrow\left(x-5\right)^4\left(x-5+1\right)\left(x-5-1\right)=0\)
\(\Rightarrow\left(x-5\right)^4\left(x-4\right)\left(x-6\right)=0\)
\(\Rightarrow x\in\left\{4;5;6\right\}\)
\(b,\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\Rightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)
\(\Rightarrow\left(2x-15\right)^3\left[\left(2xs-15\right)^2-1\right]=0\)
\(\Rightarrow\left(2x-15\right)^3\left(2x-15+1\right)\left(2x-15-1\right)=0\)
\(\Rightarrow\left(2x-15\right)^3\left(2x-14\right)\left(2x-16\right)\)
\(\Rightarrow x\in\left\{\frac{15}{2};7;8\right\}\)
Mà \(\frac{15}{2}\notin n\)
\(\Rightarrow x\in\left\{7;8\right\}\)
#)Giải :
Bài 1 :
a)\(A=\frac{2^{13}+2^5}{2^{10}+2^2}=\frac{2^5\left(2^8+1\right)}{2^2\left(2^8+1\right)}=2^3=8\)
b)\(B=\frac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}=\frac{11.3^{29}-3^{30}}{2^2.3^{28}}=\frac{11.3^{29}-3^{29}.3}{2^2.3^{28}}=\frac{3^{29}\left(11-3\right)}{2^2.3^{28}}=\frac{3^{29}.2^3}{2^2.3^{28}}=6\)
Bài 2 :
a) \(\left(x-5\right)^2=\left(x-5\right)^6\)
\(\Leftrightarrow x^4-625=x^6-15625\)
\(\Leftrightarrow x^6-x^4=15000\)
\(\Leftrightarrow x^6-x^4=5^6-5^4\)
\(\Leftrightarrow x=5\)
b)\(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\Leftrightarrow2x-15=1\)
\(\Leftrightarrow2x=16\)
\(\Leftrightarrow x=8\)