(-16-68):(-12)+(-3)^2x(-5)^3
chỉ giúp mik nhe
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a ) ( - 14 ) + 12 + ( - 12 ) . ( 23 - 7 )
= - 2 + ( - 12 ) . ( 8 - 7 )
= - 2 + ( - 12 ) . 1
= - 2 + ( - 12 )
= - 14
\(128+\left(-5\right).12+\left(-68\right):\left(-17\right)=128+\left(-60\right)+4=68+4=72\)
\(128+\left(-5\right)\cdot12+\left(-68\right):\left(-17\right)\)
\(=128+\left(-60\right)+4=68+4=72\)
\(\left(2x+1\right)\left(x^2-x\right)+x\left(5+x-2x^2\right)=3x+7\)
\(2x^3-2x^2+x^2-x+5x+x^2-2x^3=3x+7\)
\(5x-x=3x+7\)
\(4x-3x=7\)
\(x=7\)
(2x+1)(x^2-x)+x(-2x^2+x+5)=3x+7
=>2x^3-2x^2+x^2-x-2x^3+x^2+5x=3x+7
=>-x^2-x+x^2+5x=3x+7
=>4x=3x+7
=>x=7
a) (x + 5)(2x - 4) = 0
\(\Rightarrow\orbr{\begin{cases}x+5=0\\2x-4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-5\\x=2\end{cases}}}\)
b) 2(x + 5) - 3(x - 7) = 4
2x + 10 - (3x - 21) = 4
2x + 10 - 3x + 21 = 4
(-x) + 31 = 4
(-x) = 4 - 31 = -27
=> x = 27
c) (x - 4)(2x2 + 3) = 0
\(\Rightarrow\orbr{\begin{cases}x-4=0\\2x^2+3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=4\\x^2=\frac{-3}{2}\end{cases}}}\)
Vì x2 \(\ge\)0
Mà -3/2 < 0
=> Không có giá trị thõa mãn ở trường hợp x2
Vậy x = 4
a) => (12x-4)+(6x+15)=16
=>12x-48+6x+15=16
=>(12x+6x)-(15-48)=16
=>18x-(-33)=16
=>18x=16+(-33)
=>18x=-17
=>x=-17/18
KL
b) =>(-20x+(-40))-(12-30x)=48
=>-20x+(-40)-12+30x=48
=>(-20x+30x)+(-40-12)=48
=>10x+(-52)=48
=>10x=48-(-52)
=>10x=4
=>x=4/10
KL
a) 12( x - 4 ) + 3( 2x + 5 ) = 16
<=> 12x - 48 + 6x + 15 = 16
<=> 18x - 33 = 16
<=> 18x = 49
<=> x = 49/18
b) -10( 2x + 4 ) - 2( 6 - 15x ) = 48
<=> -20x - 40 - 12 + 30x = 48
<=> 10x - 52 = 48
<=> 10x = 100
<=> x = 10
\(\dfrac{2}{5}+\dfrac{1}{3}=\dfrac{6+5}{15}=\dfrac{11}{15}\)
\(\dfrac{16}{24}+\dfrac{1}{3}=\dfrac{2}{3}+\dfrac{1}{3}=1\)
\(\dfrac{7}{12}+\dfrac{4}{6}+\dfrac{3}{8}=\dfrac{14}{24}+\dfrac{16}{24}+\dfrac{9}{24}=\dfrac{39}{24}=\dfrac{13}{8}\)
\(1+\dfrac{1}{12}=\dfrac{12+1}{12}=\dfrac{13}{12}\)
\(a,-\dfrac{12}{16}-\left(\dfrac{3}{4}-x\right)=-\dfrac{5}{3}\)
\(\dfrac{3}{4}-x=-\dfrac{12}{16}-\left(-\dfrac{5}{3}\right)\)
\(\dfrac{3}{4}-x=\dfrac{11}{12}\)
\(x=\dfrac{3}{4}-\dfrac{11}{12}\)
\(x=-\dfrac{1}{6}\)
\(b,x-\dfrac{3}{7}:\dfrac{9}{14}=-\dfrac{7}{3}\)
\(x-\dfrac{3}{7}=-\dfrac{7}{3}\times\dfrac{9}{14}\)
\(x-\dfrac{3}{7}=-\dfrac{3}{2}\)
\(x=-\dfrac{3}{2}+\dfrac{3}{7}\)
\(x=-\dfrac{15}{14}\)
\(c,-\dfrac{3}{4}x+\dfrac{5}{8}x=\dfrac{1}{3}\)
\(\left(-\dfrac{3}{4}+\dfrac{5}{8}\right)x=\dfrac{1}{3}\)
\(-\dfrac{1}{8}x=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}:\left(-\dfrac{1}{8}\right)\)
\(x=-\dfrac{8}{3}\)