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a) \(\left(x+1\right)\left(x+2\right)=272\)
\(\Rightarrow x^2+3x+2=272\)
\(\Rightarrow x^2+3x-270=0\)
\(\Rightarrow x^2+18x-15x-270=0\)
\(\Rightarrow x\left(x+18\right)-15\left(x+18\right)=0\)
\(\Rightarrow\left(x+18\right)\left(x-15\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+18=0\\x-15=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-18\\x=15\end{matrix}\right.\)
d) \(\left(x+4\right)\left(x+5\right)=552\)
\(\Rightarrow x^2+9x+20=552\)
\(\Rightarrow x^2+9x-532=0\)
\(\Rightarrow x^2+28x-19x-532=0\)
\(\Rightarrow x\left(x+28\right)-19\left(x+28\right)=0\)
\(\Rightarrow\left(x+28\right)\left(x-19\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+28=0\\x-19=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-28\\x=19\end{matrix}\right.\)
`#040911`
`a)`
`3 1/3 x + 16 3/4 = -13,25`
`=> 3 1/3 x = -13,25 - 16 3/4`
`=> 3 1/3 x = -30`
`=> x = -30 \div 3 1/3`
`=> x =-9`
Vậy, `x = -9`
`b)`
`3 2/7*x - 1/8 = 2 3/4`
`=> 3 2/7x = 2 3/4 + 1/8`
`=> 3 2/7x = 23/8`
`=> x = 23/8 \div 3 2/7`
`=> x = 7/8`
Vậy, `x = 7/8`
`c)`
`x \div 4 1/3 = -2,5`
`=> x = -2,5 * 4 1/3`
`=> x = -65/6`
Vậy, `x = -65/6`
`d)`
`( (3x)/7 + 1) \div (-4) = (-1)/28`
`=> (3x)/7 +1 = (-1)/28 * (-4)`
`=> (3x)/7 + 1 = 1/7`
`=> (3x)/7 = 1/7 - 1`
`=> (3x)/7 = -6/7`
`=> 3x = -6`
`=> x = -6 \div 3`
`=> x = -2`
Vậy, `x = -2.`
a
=>10/3 . x + 16 + 3/4 = -13,25
=>10/3 x + 3/4 = -29,25
=>10/3 x = -30
=>x=-30 : 10/3
=>x=-30 . 3/10
=>x=-9
b.
=>23/7 x - 1/8 = = 11/4
=>23/7 x = 11/4 + 1/8
=>23/7 x= 22/8 + 1/8
=>23/7 x= 23/8
=>x=23/8 : 23/7
=>x=23/8 . 7/23
=>x=7/8
c.
=>x : 13/3 =-5/2
=>x=-5/2 . 13/3
=>x=-65/6
d.
=>3x/7 +1 = (-1/28) . (-4)
=>3x/7 + 1 = 1/7
=>3x/7 = -6/7
=>3x=-6
=>x=-2
\(a,2.\left(x-5\right)-3.\left(x+7\right)=14\)
\(2x-10-3x-21=14\)
\(-x-31=14\)
\(x=-31-14\)
\(x=-45\)
\(b,5.\left(x-6\right)-2\left(x+3\right)=12\)
\(5x-30-2x-6=12\)
\(3x-36=12\)
\(3x=12+36\)
\(3x=48\)
\(x=16\)
\(c,-5.\left(2-x\right)+4.\left(x-3\right)=10.x-15\)
\(-10+5x+4x-12=10x-15\)
\(-6x-22=10x-15\)
\(-6x-10x=-15+22\)
\(-16x=7\)
\(x=-\frac{7}{16}\)
Câu d , e f tương tự nha
\(\dfrac{4}{x}=\dfrac{y}{21}=\dfrac{28}{49}=\dfrac{28:7}{49:7}=\dfrac{4}{9}\\ Vậy:x=\dfrac{4.9}{4}=9\\ y=\dfrac{4.21}{9}=\dfrac{28}{3}\)
\(\dfrac{x}{2}=\dfrac{3}{y}\\ \Leftrightarrow x.y=2.3=6\\ Vậy:\left[{}\begin{matrix}\left(x;y\right)=\left(1;6\right)=\left(6;1\right)\\\left(x;y\right)=\left(2;3\right)=\left(3;2\right)\end{matrix}\right.\)
\(a,\dfrac{3}{4}x-\dfrac{7}{12}=\dfrac{5}{6}-\dfrac{2}{3}\\ \Rightarrow\dfrac{3}{4}x-\dfrac{7}{12}=\dfrac{1}{6}\\ \Rightarrow\dfrac{3}{4}x=\dfrac{1}{6}+\dfrac{7}{12}\\ \Rightarrow\dfrac{3}{4}x=\dfrac{3}{4}\\ \Rightarrow x=\dfrac{3}{4}:\dfrac{3}{4}\\ \Rightarrow x=1\\ b,\dfrac{-5}{x}=\dfrac{20}{28}\\ \Rightarrow\dfrac{-5}{x}=\dfrac{5}{7}\\ \Rightarrow\dfrac{-5}{x}=\dfrac{-5}{-7}\\ \Rightarrow x=-7\\ c,2\dfrac{1}{3}:x=7\\ \Rightarrow\dfrac{7}{3}:x=7\\ \Rightarrow x=\dfrac{7}{3}:7\\ \Rightarrow x=\dfrac{1}{3}\)
\(d,\dfrac{-105}{12}< x< \dfrac{20}{7}\Rightarrow x\in\left\{-8;-7;...;2\right\}\)
a: \(\Leftrightarrow x\cdot\dfrac{3}{4}=\dfrac{3}{4}\)
hay x=1
b: \(\Leftrightarrow x=\dfrac{-28\cdot5}{20}=-7\)
c: \(\Leftrightarrow x=\dfrac{7}{3}:7=\dfrac{1}{3}\)
d: \(\Leftrightarrow-8< x< 3\)
hay \(x\in\left\{-7;-6;-5;-4;-3;-2;-1;0;1;2\right\}\)
a) <=>5x-x=57+7
<=> 4x=64
<=> x=16
b) <=>4x=28+12
<=> x=40/4
<=>10
c) th1 x=0
th2 x-3=0
<=>x=3
d) th1 2x-4=0
<=> x=2
th2 x+1=0
<=> x=-1
câu e mình ko biết làm hihi
a, 5x - 7 = x + 57
5x - x = 57 + 7
4x = 64
x = 64 : 4
x = 16
b, x + 3x - 12 = 28
4x = 28 + 12
4x = 40
x = 10
c, x(x-3) = 0
\(\Rightarrow\orbr{\begin{cases}x=0\\x-3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}}\)
d, (2x-4)(x+1)=0
\(\Rightarrow\orbr{\begin{cases}2x-4=0\\x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}}\)
e, 1+2+...+x=120
=> \(\frac{x\left(x+1\right)}{2}=120\)
=> x(x+1) = 120.2
=>x(x+1) = 240
=> x(x+1) = 15.16
=> x=16