(x-5) x ( 1995 x 196 + 1997 x 1996) = 0

x-5 = 0 : ( 1995 x 196 + 1997 x 1996)

x-5 = 0

x= 0+5

x= 5

\(a,\Leftrightarrow\left(2x-3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-2\end{matrix}\right.\\ b,\Leftrightarrow x^3-27-x^3+4x=1\\ \Leftrightarrow4x=28\Leftrightarrow x=7\\ c,\Leftrightarrow4x^2-4x-8=0\\ \Leftrightarrow x^2-x-2=0\\ \Leftrightarrow\left(x-2\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\\ d,\Leftrightarrow2x^2+6x+x+3=0\\ \Leftrightarrow\left(x+3\right)\left(2x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{1}{2}\end{matrix}\right.\)

k) 4 x 113 x 25 - 5 x 112 x 20

= (4 x 25) x 113 - (5 x 20) x 112

= 100 x 113 - 100 x 112

= 100 x (113 - 112)

= 100

p) 1235 x 6789 x (630 - 315 x 2) : 1996

= 1235 x 6789 x (630 - 630) : 1996

= 1235 x 6789 x 0 : 1996

= 0

l) 54 x 113 + 45 x 113 + 113

= 113 (54 + 45 + 1)

= 113 x 100

= 11300

u) (500 x 9 - 250 x 18) x (1 + 2 + 3 + ... + 9)

= (2 x 250 x 9 - 250 x 2 x 9) x (1 +2 + 3 + ... + 9)

= 0 x (1 + 2 + 3 + ... + 9)

= 0

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Mình xin lỗi câu này mình ko biết

a) \(326-2x=78\Rightarrow2x=326-78=248\)

\(\Rightarrow x=248:2=124\)

b) \(42.5-\left(3x+6\right)=3^4\Rightarrow210-\left(3x+6\right)=81\)

\(\Rightarrow3x+6=210-81=129\Rightarrow3x=129-6=123\)

\(\Rightarrow x=123:3=41\)

a, 326 -2x =78

\(\Rightarrow\)2x=326-78=248

\(\Rightarrow\)x=248/2=124

b,\(42.5-\left(3x+6\right)=3^4\)

\(\Rightarrow\)\(210-3x-6=81\)

\(\Rightarrow\)3x=123

\(\Rightarrow\)x=123/3=41

c, \(4^x:64=4^{250}\)

\(\Rightarrow\)\(4^x:4^3=4^{250}\)

\(\Rightarrow\)x=253

d, \(30x+1+2+3+...+30=495\)

\(\Rightarrow\)\(30x+465=495\)

\(\Rightarrow\)\(x=1\)

a: \(x\cdot3\cdot5=2,7\)

=>\(x\cdot15=2,7\)

=>\(x=\dfrac{2,7}{15}=0,18\)

b: \(1,2:x\cdot4=20\)

=>\(1,2:x=20:4=5\)

=>\(x=1,2:5=0,24\)

a: \(x\left(x+7\right)-\left(x-2\right)\left(x+3\right)=0\)

\(\Leftrightarrow x^2+7x-x^2-x+6=0\)

hay x=-1

b: Ta có: \(\left(x+2\right)^2-\left(x^2-4\right)=0\)

\(\Leftrightarrow x+2=0\)

hay x=-2

b. (x + 2)² - x² + 4 = 0

<=> (x + 2 - x)(x + 2 + x) + 4 = 0

<=> 2(2 + 2x) + 4 = 0

<=> 4(1 + x) + 4 = 0

<=> 4(1 + x) = -4

<=> 1 + x = -1

<=> x = -1 - 1

<=> x = -2

\(a,\) \(x\left(x+7\right)-\left(x-2\right)\left(x+3\right)\)

\(\Leftrightarrow x^2+7x-x^2-3x+2x+6\\ \Leftrightarrow6x=0\\ \Leftrightarrow x=0\)

    \(Vậy...\)

\(b,\) \(\left(x+2\right)^2-\left(x^2-4\right)=0\)

\(\Leftrightarrow x^2+4x+4-x^2+4=0\\ \Leftrightarrow4x+8=0\\ \Leftrightarrow x=-2\)

a)=> x² + 6x + 9 - x² + 4x = 39 => 10x = 30 => x = 3
b) => x(x - 9) + 2(x -9) = 0 => (x+2)(x-9) = 0 
+)th1: x + 2 = 0 => x = -2
+)th2: x - 9 =0 => x = 9 

\(a,\Rightarrow x^2+4x+25-x^2=3\\ \Rightarrow4x=-22\Rightarrow x=-\dfrac{11}{2}\\ b,\Rightarrow\left(2x-3-4x-3\right)\left(2x-3+4x+3\right)=0\\ \Rightarrow6x\left(-2x-6\right)=0\Rightarrow\left[{}\begin{matrix}x=-3\\x=0\end{matrix}\right.\)