Tim x e N
2x-7 =20+5.(-3)
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a) <=> 15-5x-20=-12-3
<=> -5x=-12-3-15+20=-10
=>x=-10:(-5)=2
b)<=>7-x-25-7=-25
<=> -x=-25-7+25+7=0 =>x=0
c) /x+2/=0 => x+2=0 =>x=-2
d) sai đề
e)<=> /x-5/ = 7
<=> \(\orbr{\begin{cases}x-5=7\\x-5=-7\end{cases}}\)
<=> \(\orbr{\begin{cases}x=12\\x=-2\end{cases}}\)
g) <=> -x-20-8+2x=-15
<=> x=-15+20+8=13
Câu a:
A = 25.2^3 - (15 - 18) + (12 - 19 + 10)
A = 25.8 + 3 + (- 7 + 10)
A = 200 + 3+ 3
A = 203 + 3
A = 206
\(\left(\frac{3}{5}-x\right)+\frac{13}{20}=\frac{5}{6}\)
\(\frac{3}{5}-x=\frac{5}{6}-\frac{13}{20}\)
\(\frac{3}{5}-x=\frac{11}{60}\)
\(x=\frac{3}{5}-\frac{11}{60}\)
\(x=\frac{5}{12}\)
\(\left(x-\frac{1}{2}\right)\cdot\frac{5}{3}=\frac{7}{4}-\frac{1}{2}\)
\(\left(x-\frac{1}{2}\right)\cdot\frac{5}{3}=\frac{5}{4}\)
\(x-\frac{1}{2}=\frac{5}{4}:\frac{5}{3}\)
\(x-\frac{1}{2}=\frac{3}{4}\)
\(x=\frac{3}{4}+\frac{1}{2}\)
\(x=\frac{5}{4}\)
( 3/5 - x ) + 13/20 = 5/6
3/5 - x = 5/6 - 13/20
3/5 - x = 89/60
x = 3/5 - 89/60
x = 25/12
( x - 1/2 ) x 5/3 = 7/4 - 1/2
( x - 1/2 ) x 5/3 = 5/4
x - 1/2 = 5/4 : 5/3
x - 1/2 = 35/12
x = 35/12 + 1/2
x = 41/12
Trả lời:
A = ( 2x - 7 )4
Ta có: \(\left(2x-7\right)^4\ge0\forall x\)
Dấu "=" xảy ra khi 2x - 7 = 0 <=> 2x = 7 <=> x = 7/2
Vậy GTNN của A = 0 khi x = 7/2
B = ( x + 1 )10 + ( y - 2 )20 + 7
Ta có: \(\left(x+1\right)^{10}\ge0\forall x;\left(y-2\right)^{20}\ge0\forall y\)
\(\Leftrightarrow\left(x+1\right)^{10}+\left(y-2\right)^{20}\ge0\forall x;y\)
\(\Leftrightarrow\left(x+1\right)^{10}+\left(y-2\right)^{20}+7\ge7\forall x;y\)
Dấu "=" xảy ra khi x + 1 = 0 <=> x = -1 và y - 2 = 0 <=> y = 2
Vậy GTNN của B = 7 khi x = -1 và y = 2
C = ( 3x - 4 )100 + ( 5y + 1 )50 - 20
Ta có: \(\left(3x-4\right)^{100}\ge0\forall x;\left(5y+1\right)^{50}\ge0\forall y\)
\(\Leftrightarrow\left(3x-4\right)^{100}+\left(5y+1\right)^{50}\ge0\forall x;y\)
\(\Leftrightarrow\left(3x-4\right)^{100}+\left(5y+1\right)^{50}-20\ge-20\forall x;y\)
Dấu "=" xảy ra khi 3x - 4 = 0 <=> x = 4/3 và 5y + 1 = 0 <=> y = -1/5
Vậy GTNN của C = -20 khi x = 4/3 và y = -1/5
D = ( 2x + 3 )20 + ( 3y - 4 )10 + 1000
Ta có: \(\left(2x+3\right)^{20}\ge0\forall x;\left(3y-4\right)^{10}\ge0\forall y\)
\(\Leftrightarrow\left(2x+3\right)^{20}+\left(3y-4\right)^{10}\ge0\forall x;y\)
\(\Leftrightarrow\left(2x+3\right)^{20}+\left(3y-4\right)^{10}+100^0\ge1\forall x;y\)
Dấu "=" xảy ra khi 2x + 3 = 0 <=> x = -3/2 và 3y - 4 = 0 <=> y = 4/3
Vậy GTNN của D = 1 khi x = -3/2 và y = 4/3
E = ( x - y )50 + ( y - 2 )60 + 3
Ta có: \(\left(x-y\right)^{50}\ge0\forall x;y\); \(\left(y-2\right)^{60}\ge0\forall y\)
\(\Leftrightarrow\left(x-y\right)^{50}+\left(y-2\right)^{60}\ge0\forall x;y\)
\(\Leftrightarrow\left(x-y\right)^{50}+\left(y-2\right)^{60}+3\ge3\forall x;y\)
Dấu "=" xảy ra khi x - y = 0 <=> x = y và y - 2 = 0 <=> y = 2
Vậy GTNN của E = 3 khi x = y = 2
a ) 2 - x = 5 - 6
2 - x = - 1
x = 2 - ( - 1 )
x = 3
Vậy x = 3
b ) 7 - x = 17 - ( - 5 )
7 - x = 22
x = 7 - 22
x = - 15
Vậy x = - 15
Ta có :
a) 2 - x = 5 - 6
=> 2 - x = - 1
=> x = 3
b ) 7 - x = 17 - ( - 5 )
=> 7 - x = 22
=> x = - 15
c ) | x + 3 | = 7
=> x + 3 = 7 hoặc x + 3 = - 7
=> x = 4 hoặc x = - 10
a) \(\dfrac{13}{20}+\dfrac{3}{5}+x=\dfrac{5}{6}\)
\(\Rightarrow\dfrac{5}{4}+x=\dfrac{5}{6}\)
\(\Rightarrow x=\dfrac{5}{6}-\dfrac{5}{4}\)
\(\Rightarrow x=\dfrac{-5}{12}\)
b) \(x+\dfrac{1}{3}=\dfrac{2}{5}-\dfrac{-1}{3}\)
\(\Rightarrow x+\dfrac{1}{3}=\dfrac{11}{15}\)
\(\Rightarrow x=\dfrac{11}{15}-\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{2}{5}\)
c)\(\dfrac{-5}{8}-x=\dfrac{-3}{20}-\dfrac{-1}{6}\)
\(\dfrac{-5}{8}-x=\dfrac{1}{60}\)
\(\Rightarrow x=\dfrac{-5}{8}-\dfrac{1}{60}\)
\(\Rightarrow x=\dfrac{-77}{120}\)
d) \(\dfrac{3}{5}-x=\dfrac{1}{4}+\dfrac{7}{10}\)
\(\Rightarrow\dfrac{3}{5}-x=\dfrac{19}{20}\)
\(\Rightarrow x=\dfrac{3}{5}-\dfrac{19}{20}\)
\(\Rightarrow x=\dfrac{-7}{20}\)
e) \(\dfrac{-3}{7}-x=\dfrac{4}{5}+\dfrac{-2}{3}\)
\(\Rightarrow\dfrac{-3}{7}-x=\dfrac{2}{15}\)
\(\Rightarrow x=\dfrac{-3}{7}-\dfrac{2}{15}\)
\(\Rightarrow x=\dfrac{-59}{105}\)
g) \(\dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)
\(\Rightarrow\dfrac{-5}{6}-x=\dfrac{1}{4}\)
\(\Rightarrow x=\dfrac{-5}{6}-\dfrac{1}{4}\)
\(\Rightarrow x=\dfrac{-13}{12}\)
a) 12 : x = 3
X = 12 : 3
X = 4
b) 21 : x = 7
X = 21 : 7
X = 3
c) 30 : x = 3
X = 30 : 3
X = 10
d) X : 7 = 4
X = 4 x 7
X = 28
e) 20 : x = 5
X = 20 : 5
X = 4
f) X x 6 = 42
X = 42 : 6
X = 7
2x - 7 = 20 + 5.(-3)
=> 2x - 7 = 20 - 15
=> 2x = 5 + 7
=> 2x = 12
=> x = 12 : 2
=> x = 6
\(2x-7=20+5.\left(-3\right)\)
\(\Leftrightarrow2x-7=20+\left(-15\right)\)
\(\Leftrightarrow2x-7=5\)
\(\Leftrightarrow2x=12\Leftrightarrow x=6\)