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a)
\(\frac{x+5}{4}-\frac{2x-3}{3}=\frac{6x-1}{8}+\frac{2x-1}{12}\\ \Leftrightarrow\frac{6x+30}{24}-\frac{16x-24}{24}-\frac{18x-3}{24}-\frac{4x-2}{24}=0\\ \Leftrightarrow\frac{6x+30-16x+24-18x+3-4x+2}{24}=0\\ \Leftrightarrow\frac{59-32x}{24}=0\\ \Rightarrow59-32x=0\\ \Rightarrow x=\frac{59}{32}\)
b)
\(\frac{x+4}{5}-x+4=\frac{x}{3}-\frac{x-2}{2}\\ \Leftrightarrow\frac{6x+24-30x+120-10x+15x-30}{30}=0\\ \Leftrightarrow\frac{114-19x}{30}=0\\ \Rightarrow114-19x=0\\ \Rightarrow x=\frac{-144}{-19}=6\\ \Rightarrow x=6\)
c)
\(x^2-3x+2=0\\ \Leftrightarrow2-x-2x+x^2=0\\ \Leftrightarrow2\cdot\left(1-x\right)-x\cdot\left(1-x\right)=0\\ \Leftrightarrow\left(2-x\right)\cdot\left(1-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}2-x=0\\1-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\x=1\end{matrix}\right.\)
a. \(1+\frac{2x-5}{6}=\frac{3-x}{4}\)
\(\Leftrightarrow\frac{12}{12}+\frac{2\left(2x-5\right)}{12}-\frac{3\left(3-x\right)}{12}=0\)
\(\Leftrightarrow12+4x-10-9+3x=0\)
\(\Leftrightarrow7x-7=0\)
\(\Leftrightarrow7x=7\Leftrightarrow x=1\)
b. \(\frac{x+1}{2}-\frac{x-2}{3}=\frac{3\left(x+1\right)}{6}-\frac{2\left(x-2\right)}{6}=\frac{x+7}{6}\)
c. \(\frac{2x-1}{3}+x=\frac{x+4}{2}\)
\(\Leftrightarrow\frac{2\left(2x-1\right)}{6}+\frac{6x}{6}-\frac{3\left(x+4\right)}{6}=0\)
\(\Leftrightarrow7x-14=0\)
\(\Leftrightarrow7x=14\Leftrightarrow x=2\)
d. \(\frac{x+5}{4}-\frac{2x-3}{3}-\frac{6x-1}{8}+\frac{2x-1}{12}\)
\(=\frac{6\left(x+5\right)}{24}-\frac{8\left(2x-3\right)}{24}-\frac{3\left(6x-1\right)}{24}+\frac{2\left(2x-1\right)}{24}\)
\(=6x+30-16x+24-18x+3+4x-2\)
\(=-24x-55\)
Mình ko ghi lại đề , bạn ghi ra xong rồi suy ra như mình nha .
1) \(=>A=\left(6x^2+3x-10x-5\right)-\left(6x^2+14x-9x-21\right)\)
\(=>A=-12x+16\)
2) \(=>B=8x^3+27-8x^3+2=29\)
3)\(=>C=[\left(x-1\right)-\left(x+1\right)]^3=\left(-2\right)^3=-8\)
4)\(=>D=[\left(2x+5\right)-\left(2x\right)]^3=5^3=125\)
5)\(=>E=\left(3x+1\right)^2-\left(3x+5\right)^2+12x+2\left(6x+3\right)\)
\(=>E=\left(3x+1+3x+5\right)\left(3x+1-3x-5\right)+12x+12x+6\)
\(=>E=\left(6x+6\right)\left(-4\right)+24x+6=-24x-24+24x+6=-18\)
6)\(=>F=\left(2x^2+3x-10x-15\right)-\left(2x^2-6x\right)+x+7=-8\)
k cho mik nha ,
a) 4( 18 - 5x ) - 12( 3x - 16 ) = 15( 2x - 16 ) - 6( x + 14 )
<=> 72 - 20x - 36x + 192 = 30x - 240 - 6x - 84
<=> -20x - 36x - 30x + 6x = -240 - 84 - 72 - 192
<=> -80x = -588
<=> x = -588/-80 = 147/20
b) ( x + 3 )( x + 2 ) - ( x - 2 )( x + 5 ) = 6
<=> x2 + 5x + 6 - ( x2 + 3x - 10 ) = 6
<=> x2 + 5x + 6 - x2 - 3x + 10 = 6
<=> 2x + 16 = 6
<=> 2x = -10
<=> x = -5
c) -x( x + 3 ) + 2 = ( 4x + 1 )( x - 1 ) + 2x
<=> -x2 - 3x + 2 = 4x2 - 3x - 1 + 2x
<=> -x2 - 3x - 4x2 + 3x - 2x = -1 - 2
<=> -5x2 - 2x = -3
<=> -5x2 - 2x + 3 = 0
<=> -( 5x2 + 2x - 3 ) = 0
<=> -( 5x2 + 5x - 3x - 3 ) = 0
<=> -[ 5x( x + 1 ) - 3( x + 1 ) ] = 0
<=> -( x + 1 )( 5x - 3 ) = 0
<=> \(\orbr{\begin{cases}x+1=0\\5x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\frac{3}{5}\end{cases}}\)
d) ( 2x + 3 )( x - 3 ) - ( x - 3 )( x + 1 ) = ( 2 - x )( 3x + 1 ) + 3
<=> 2x2 - 3x - 9 - ( x2 - 2x - 3 ) = -3x2 + 5x + 2 + 3
<=> 2x2 - 3x - 9 - x2 + 2x + 3 = -3x2 + 5x + 2 + 3
<=> 2x2 - 3x - x2 + 2x + 3x2 - 5x = 2 + 3 + 9 - 3
<=> 4x2 - 6x = 11
<=> 4x2 - 6x - 11 = 0
=> Vô nghiệm ( Lớp 8 chưa học nghiệm vô tỉ nên để vậy ) :))
vẫn làm được nha quỳnh !
\(4x^2-6x-11=0\)
\(< =>\left(4x^2-6x+\frac{9}{4}\right)-13\frac{1}{4}=0\)
\(< =>\left(2x-\frac{3}{2}\right)^2=\frac{53}{4}\)
\(< =>\orbr{\begin{cases}2x-\frac{3}{2}=\frac{\sqrt{53}}{2}\\2x-\frac{3}{2}=-\frac{\sqrt{53}}{2}\end{cases}}\)
\(< =>\orbr{\begin{cases}2x=\frac{3+\sqrt{53}}{2}\\2x=\frac{3-\sqrt{53}}{2}\end{cases}}\)
\(< =>\orbr{\begin{cases}x=\frac{3+\sqrt{53}}{4}\\x=\frac{3-\sqrt{53}}{4}\end{cases}}\)
f/ \(3xy\left(x+y\right)-\left(x+y\right)\left(x^2+y^2+2xy\right)+y^3=27\)
\(3x^2y+3xy^2-\left(x+y\right)\left(x+y\right)^2+y^3=27\)
\(3x^2y+3xy^3-\left(x+y\right)^3+y^3=27\)
\(3x^2y+3xy^3-\left(x^3+3x^2y+3xy^2+b^3\right)+y^3=27\)
\(-x^3=27\)
\(x=-3\)
Bài 1 :
a) \(3x\left(5x^2-2x-1\right)=3x\cdot5x^2+3x\left(-2x\right)+3x\left(-1\right)\)
\(=15x^3-6x^2-3x\)
b) \(\left(x^2-2xy+3\right)\left(-xy\right)\)
\(=x^2\left(-xy\right)-2xy\left(-xy\right)+3\left(-xy\right)\)
\(=-x^3y+2x^2y^2-3xy\)
c) \(\frac{1}{2}x^2y\left(2x^3-\frac{2}{5}xy-1\right)\)
\(=\frac{1}{2}x^2y\cdot2x^3+\frac{1}{2}x^2y\cdot\left(-\frac{2}{5}xy\right)+\frac{1}{2}x^2y\left(-1\right)\)
\(=x^5y-\frac{1}{5}x^3y^2-\frac{1}{2}x^2y\)
d) \(\frac{1}{2}xy\left(\frac{2}{3}x^2-\frac{3}{4}xy+\frac{4}{5}y^2\right)\)
\(=\frac{1}{2}xy\cdot\frac{2}{3}x^2+\frac{1}{2}xy\cdot\left(-\frac{3}{4}xy\right)+\frac{1}{2}xy\cdot\frac{4}{5}y^2\)
\(=\frac{1}{3}x^3y-\frac{3}{8}x^2y^2+\frac{2}{5}xy^3\)
e) \(\left(x^2y-xy+xy^2+y^3\right)\left(3xy^3\right)\)
= \(x^2y\cdot3xy^3-xy\cdot3xy^3+xy^2\cdot3xy^3+y^3\cdot3xy^3\)
\(=3x^3y^4-3x^2y^4+3x^2y^5+3xy^6\)
Bài 2 :
3(2x - 1) + 3(5 - x) = 6x - 3 + 15 - x = (6x - x) - 3 + 15 = 5x - 3 + 15
Thay x = -3/2 vào biểu thức trên ta có : \(5\cdot\left(-\frac{3}{2}\right)-3+15\)
\(=-\frac{15}{2}-3+15=\frac{9}{2}\)
b) 25x - 4(3x - 1) + 7(5 - 2x)
= 25x - 12x + 4 + 35 - 14x
= (25x - 12x - 14x) + 4 + 35 = -x + 4 + 35 = -x + 39
Thay \(x=2\)vào biểu thức trên ta có : -2 + 39 = 37
c) 4x - 2(10x + 1) + 8(x - 2)
= 4x - 20x - 2 + 8x - 16
= (4x - 20x + 8x) - 2 - 16 = -8x - 2 - 16 = -8x - 18
Thay x = 1/2 vào biểu thức trên ta có \(-8\cdot\frac{1}{2}-18=-4-18=-22\)
d) Tương tự
Bài 3:
a) \(2x\left(x-4\right)-x\left(2x+3\right)=4\)
=> 2x2 - 8x - 2x2 - 3x = 4
=> (2x2 - 2x2) + (-8x - 3x) = 4
=> -11x = 4
=> x = \(-\frac{4}{11}\)
b) x(5 - 2x) + 2x(x - 7) = 18
=> 5x - 2x2 + 2x2 - 14x = 18
=> 5x - 14x = 18
=> -9x = 18
=> x = -2
Còn 2 câu làm tương tự
Bạn chép lại đề đi ạ. Dùng biểu tượng \(\Sigma\) để gõ câu hỏi nhé.