3\(x\) - 28  = \(x\) + 36

3\(x\) - \(x\)    = 36 + 28

2\(x\)         = 64

   \(x\)         = 64 : 2

   \(x\)         = 32

3x - 28 = x+36

3x - x = 36+28

2x      =64

x=64:2

x=32

a) \(-45:5.\left(-3-2x\right)=3\)

\(-9.\left(-3-2x\right)=3\)

\(-3-2x=\left(3:-9\right)\)

\(-3-2x=\dfrac{-1}{3}\)

\(-2x=-3-\dfrac{1}{3}\)

-2x=\(\dfrac{-10}{3}\)

\(x=\dfrac{-10}{3}:-2\)

\(x=\dfrac{5}{3}\)

b) 

3x - 28 = x + 36

<=> 3x - x = 36 + 28

<=> 2x = 64

<=> x = 32

Vậy x = 32

c) 

(-12)2.x = 56 + 10.13.x

144.x = 56 + 130.x

144x – 130 x = 56

14x = 56

x = 56: 14

x = 4

Vậy x = 4

\(a,-45:5\cdot\left(-3-2x\right)=3\\ \Leftrightarrow-9\cdot\left(-3-2x\right)=3\\ \Leftrightarrow-2x-3=-\dfrac{1}{3}\\ \Leftrightarrow2x+3=\dfrac{1}{3}\\ \Leftrightarrow2x=-\dfrac{8}{3}\\ \Leftrightarrow x=-\dfrac{4}{3}\\ b,3x-28=x+36\\ \Leftrightarrow2x=64\\ \Leftrightarrow x=32\\ c,\left(-12\right)^2x=56+10\cdot13x\\ \Leftrightarrow144x=56+130x\\ \Leftrightarrow14x=56\\ \Leftrightarrow x=4\)

a.

\(10-2\left(4-3x\right)=-4\)

\(\Leftrightarrow2\left(4-3x\right)=10+4\)

\(\Leftrightarrow2\left(4-3x\right)=14\)

\(\Leftrightarrow4-3x=7\)

\(\Leftrightarrow3x=-3\)

\(\Leftrightarrow x=-1\)

b.

\(-12+3\left(-x+7\right)=-18\)

\(\Leftrightarrow3\left(-x+7\right)=-18+12=-6\)

\(\Leftrightarrow-x+7=-6:3=-2\)

\(\Leftrightarrow x=9\)

c.

\(-45:5.\left(-3-2x\right)=3\)

\(\Leftrightarrow-9.\left(-3-2x\right)=3\)

\(\Leftrightarrow-3-2x=-\dfrac{1}{3}\)

\(\Leftrightarrow2x=-\dfrac{8}{3}\)

\(\Leftrightarrow x=-\dfrac{4}{3}\notin Z\left(loại\right)\)

Câu này em ghi sai đề?

d.

\(3x-28=x+36\)

\(\Leftrightarrow2x=28+36\)

\(\Leftrightarrow2x=64\)

\(\Leftrightarrow x=32\)

e.

\(\left(-12\right)^2.x=56+10.13x\)

\(\Leftrightarrow144x=56+130x\)

\(\Leftrightarrow144x-130x=56\)

\(\Leftrightarrow14x=56\)

\(\Leftrightarrow x=4\)

a: \(\Leftrightarrow\dfrac{3x-2}{\left(x-2\right)\left(x-10\right)}-\dfrac{4x+3}{\left(x+8\right)\left(x-2\right)}=\dfrac{8x+11}{\left(x-10\right)\left(x+8\right)}\)

=>(3x-2)(x+8)-(4x+3)(x-10)=(8x+11)(x-2)

=>3x^2+24x-2x-16-4x^2+40x-3x+30=8x^2-16x+11x-22

=>-x^2+59x+14-8x^2+5x+22=0

=>-9x^2+54x+36=0

=>x^2-6x-4=0

=>\(x=3\pm\sqrt{13}\)

b: \(\Leftrightarrow\dfrac{2x-5}{\left(x+9\right)\left(x-4\right)}-\dfrac{x-6}{\left(x+7\right)\left(x-4\right)}=\dfrac{x+8}{\left(x+9\right)\left(x+7\right)}\)

=>(2x-5)(x+7)-(x-6)(x+9)=(x+8)(x-4)

=>2x^2+14x-5x-35-x^2-9x+6x+54=x^2+4x-32

=>x^2+6x+19=x^2+4x-32

=>2x=-51

=>x=-51/2

Lời giải:

a) Đa thức không phân tích được thành nhân tử

b)

\((x^2-3x)^2-3(x^2-3x)+2\)

\(=a^2-3a+2\) (đặt \(x^2-3x=a)\)

\(=a^2-a-2a+2=a(a-1)-2(a-1)=(a-1)(a-2)\)

\(=(x^2-3x-1)(x^2-3x-2)\)

a) \(58+7x=100\)

\(=>7x=100-58\)

\(=>7x=42\)

\(=>x=42:7\)

\(=>x=6\)

b) \(3x-7=28\)

\(=>3x=28+7\)

\(=>3x=35\)

\(=>x=35:3\)

\(=>x=\dfrac{35}{3}\)

c) \(x-56:4=16\)

\(=>x-14=16\)

\(=>x=16+14\)

\(=>x=30\)

d) \(101+\left(36-4x\right)=105\)

\(=>36-4x=105-101\)

\(=>36-4x=4\)

\(=>4x=36-4\)

\(=>4x=32\)

\(=>x=32:4\)

\(=>x=8\)

e) \(\left(x-12\right):12=12\)

\(=>x-12=12.12\)

\(=>x-12=144\)

\(=>x=144-12\)

\(=>x=132\)

f) \(\left(3x-2^4\right).7^3=2.7^4\)

\(=>3x-2^4=2.7^4:7^3\)

\(=>3x-16=2.7=14\)

\(=>3x=14+16\)

\(=>3x=30\)

\(=>x=30:3\)

\(=>x=10\)

i) \(\left(10+2x\right).4^{2011}=4^{2013}\)

\(=>10+2x=4^{2013}:4^{2011}\)

\(=>10+2x=4^2=16\)

\(=>2x=16-10\)

\(=>2x=6\)

\(=>x=6:2\)

\(=>x=3\)

\(#WendyDang\)

a,ĐK: x≥4

Ta có: \(2\sqrt{x-4}-\dfrac{1}{3}\sqrt{9x-36}=4-\sqrt{x-4}\)

      \(\Leftrightarrow2\sqrt{x-4}-\sqrt{x-4}=4-\sqrt{x-4}\)

      \(\Leftrightarrow2\sqrt{x-4}=4\)

      \(\Leftrightarrow\sqrt{x-4}=2\Leftrightarrow x-4=4\Leftrightarrow x=8\left(tm\right)\)

b, ĐK: x≥2

Ta có: \(3\sqrt{x-2}-\sqrt{x^2-4}=0\)

      \(\Leftrightarrow3\sqrt{x-2}-\sqrt{\left(x-2\right)\left(x+2\right)}=0\)

      \(\Leftrightarrow\sqrt{x-2}\left(3-\sqrt{x+2}\right)=0\)

      \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-2}=0\\3-\sqrt{x+2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-2=0\\\sqrt{x+2}=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x+2=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=7\end{matrix}\right.\)

=36x(28+82)+64x(69+41)

=36x110+64x110

=110x(26+64)

=110x100

=11000

**** bn hiền

36x28+36x82+64x69+64x41

=36x(28+82)+64x(69+41)

=36x110+64x110

=110x(36+64)

=110x100

=11000

36 x ( 28 + 82 ) + 64 x ( 69 + 41 ) = ( 36 + 64 ) x  (28 + 82)  = 100 x 110 = 11000

\(\frac{7}{9}+\frac{5}{12}-\frac{3}{4}\)

\(=\frac{7x4}{36}+\frac{5x3}{36}-\frac{3x9}{36}\)

\(=\frac{28+15-27}{36}\)

\(=\frac{16}{36}=\frac{4}{9}\)