3 x 7 - 2 x ( X + 1 ) = 7
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\(8x-48+4x-12-14=-x+4\)
\(\Leftrightarrow12x-75=-x+4\Leftrightarrow13x=79\Leftrightarrow x=\dfrac{79}{13}\)
\(-7\left(8-x\right)-6\left(x+9\right)=20-x\Leftrightarrow-56+7x-6x-54=20-x\)
\(\Leftrightarrow2x=130\Leftrightarrow x=65\)
\(9x-63-80+60x=-7x+15\Leftrightarrow76x=158\Leftrightarrow x=\dfrac{79}{38}\)
\(-96-16x-60+30x=-40x-16\Leftrightarrow54x=140\Leftrightarrow x=\dfrac{70}{27}\)
\(17x-102-14x-28=4x-24-2x+4\Leftrightarrow x=110\)
3/25 x ( 15/7 - 2/7 ) + 3/7 x 1/25
= 3/25 x 13/7 + 3/7 x 1/25
= (3 x 13/7 + 3/7 ) x 1/25
= 42/7 x 1/25
= 6 x 1/25
= 6/25
\(\dfrac{3}{25}\times\dfrac{15}{7}+\dfrac{3}{7}\times\dfrac{1}{25}-\dfrac{2}{7}\times\dfrac{3}{25}\)
\(=\dfrac{3}{25}\times\left(\dfrac{15}{7}-\dfrac{2}{7}\right)+\dfrac{3}{7}\times\dfrac{1}{25}\)
\(=\dfrac{3}{25}\times\dfrac{13}{7}+\dfrac{3}{7}\times\dfrac{1}{25}\)
\(=\dfrac{3\times13}{25\times7}+\dfrac{3\times1}{7\times25}\)
\(=\dfrac{39}{175}+\dfrac{3}{175}\)
\(=\dfrac{39+3}{175}\)
\(=\dfrac{42}{175}\)
\(=\dfrac{6}{25}\)
Bài 1 :
Mình nghĩ phải sửa đề ntn :
\(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)
\(\Leftrightarrow\left[2\left(2x+7\right)\right]^2-\left[3\left(x+3\right)\right]^2=0\)
\(\Leftrightarrow\left[2\left(2x+7\right)-3\left(x+3\right)\right]\left[2\left(2x+7\right)+3\left(x+3\right)\right]=0\)
\(\Leftrightarrow\left(4x+14-3x-9\right)\left(4x+14+3x+9\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(7x+23\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+5=0\\7x+23=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=\frac{-23}{7}\end{cases}}}\)
Vậy....
b) \(A=\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)
Đặt \(q=x^2+x+1\)ta có :
\(A=q\left(q+1\right)-12\)
\(A=q^2+q-12\)
\(A=q^2+4q-3q-12\)
\(A=q\left(q+4\right)-3\left(q+4\right)\)
\(A=\left(q+4\right)\left(q-3\right)\)
Thay \(q=x^2+x+1\)ta có :
\(A=\left(x^2+x+1+4\right)\left(x^2+x+1-3\right)\)
\(A=\left(x^2+x+5\right)\left(x^2+x-2\right)\)
\(A=\left(x^2+x+5\right)\left(x^2+2x-x-2\right)\)
\(A=\left(x^2+x+5\right)\left[x\left(x+2\right)-\left(x+2\right)\right]\)
\(A=\left(x^2+x+5\right)\left(x+2\right)\left(x-1\right)\)
-15 ( x - 2 ) + 7 ( 3 - x ) = 7
-15x + 30 + 21 - 7x = 7
-15x - 7x = 7 - 21 - 30
-22x = -44
x = ( -44 ) : ( -22)
x = 2
Vậy x = 2
-12 ( x - 5 ) + 7 ( 3 - x ) = 5
-12x + 60 + 21 - 7x = 5
-12x - 7x = 5 - 21 - 60
-19x = -76
x = ( -76 ) : (-19)
x = 4
Vậy x = 4.
# HOK TỐT #
a) 2 + 1/3 - x = 1 + 1/4
7/3 -x = 5/4
x = 7/3 - 5/4
x = 13/12
b) (2/7 x 2) : x = 1 :7/2
4/7 : x = 2/7
x = 4/7 : 2/7
x = 2
a) 2 + 1/3 - x = 1 + 1/4
7/3 -x = 5/4
x = 7/3 - 5/4
x = 13/12
b) (2/7 x 2) : x = 1 :7/2
4/7 : x = 2/7
x = 4/7 : 2/7
x = 2
|7 - \(\dfrac{3}{4}\)\(x\)| - \(\dfrac{3}{2}\) = \(\dfrac{1}{\dfrac{1}{2}}\)
|7 - \(\dfrac{3}{4}x\)| - \(\dfrac{3}{2}\) = 2
|7 - \(\dfrac{3}{4}\)\(x\)| = 2 + \(\dfrac{3}{2}\)
|7 - \(\dfrac{3}{4}x\)| = \(\dfrac{7}{2}\)
\(\left[{}\begin{matrix}7-\dfrac{3}{4}x=\dfrac{7}{2}\\7-\dfrac{3}{4}x=-\dfrac{7}{2}\end{matrix}\right.\)
\(\left[{}\begin{matrix}\dfrac{3}{4}x=7-\dfrac{7}{2}\\\dfrac{3}{4}=7+\dfrac{7}{2}\end{matrix}\right.\)
\(\left[{}\begin{matrix}\dfrac{3}{4}x=\dfrac{7}{2}\\\dfrac{3}{4}x=\dfrac{21}{2}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{14}{3}\\x=14\end{matrix}\right.\)
5 - |\(x-3\)| = 5
|\(x-3\)| = 5 - 5
|\(x-3\)| = 0
\(x-3\) = 0
\(x\) = 3
Ta có : x2 - 2x - (x + 3)2 = 6
<=> x2 - 2x - x2 - 6x - 9 = 6
<=> -8x - 9 = 6
=> -8x = 15
=> x = \(\frac{15}{-8}\)
`3 xx 7 - 2 xx ( x +1)=7`
`21-2xx(x+1)=7`
`2xx(x+1)=21-7`
`2xx(x+1)=14`
`x+1=14:2`
`x+1=7`
`x=7-1`
`x=6`
3 x 7 - 2 x ( X + 1 ) = 7
21 - 2 x (X + 1 ) = 7
2 x X +1 = 21: 7
2 x X + 1 = 3
X x 2 + 1 = 3
X x 3 = 3
X = 3 : 3
x = 1