1084 +(942 -2x) :36 x44 =1434

=>(942 -2x) :36 x44 =1434 -1084

=>(942 -2x) :36 x44 = 350

=>(942 -2x) :36 =350 :44

=>(942 -2x) :36 =175 /22

=>(942 -2x)=175 /22x 36

=>(942 -2x)= 3150 /11

=>2x = 3150 /11-942

=>2x =-7212 /11

kết quả nè còn lại bn tự nghĩ nhé! -655,6363636

xin lôi nhưng mink bn rui k ghi dc

\(1084+\left(942-2x\right)\div36\cdot44=1434\)

\(\left(942-2x\right)\div36\cdot44=1434-1084\)

\(\left(942-2x\right)\div36\cdot44=350\)

\(\left(942-2x\right)\div36=350\div44\)

\(\left(942-2x\right)\div36=\frac{175}{22}\)

\(942-2x=\frac{175}{22}\cdot36\)

\(942-2x=\frac{3150}{11}\)

\(2x=942-\frac{3150}{11}\)

\(2x=\frac{7212}{11}\)

\(\Rightarrow x=\frac{3606}{11}\)

lời giải nè

f(x)=x¹⁴-(13+1)x¹³+(13+1)x¹²-....+(13+1)x²-(13+1)x+(13+1)

mà theo đầu bài f(x)=13 => chỗ nào có 13 ta thay thành x

=>f(13)=x¹⁴-(x+1)x¹³+(x+1)x¹³-.......+(x+1)x²-(x+1)x+(x+1)

<=>f(13)=x¹⁴-x¹⁴-x¹³+x¹⁴+x¹³-.......+x^3_x²-x²-x+x+1=1

=>f(13)=1

k cho mk nha!!!

a) \(\left(19x-2.52\right):14=\left(13-8\right).2-42\)

\(\left(19x-50\right):14=10-42\)

\(\left(19x-50\right):14=-32\)

\(19x-50=-448\)

\(19x=-398\)

\(x=-\dfrac{398}{19}\)

viết latex đi bn ơi!

a,x=6² ...6⁸

b: Ta có: \(100< 5^{2x-1}< 5^6\)

\(\Leftrightarrow2x-1\in\left\{3;5\right\}\)

hay \(x\in\left\{2;3\right\}\)

\(=\left(13x^2-13x\right)-\left(14x-14\right)=13x\left(x-1\right)-14\left(x-1\right)=\left(13x-14\right)\left(x-1\right)\)

a: 49x^2-25=0

=>(7x-5)(7x+5)=0

=>7x-5=0 hoặc 7x+5=0

=>x=5/7 hoặc x=-5/7

b: Đề thiếu vế phải rồi bạn

c: (3x-2)^2-9(x+4)(x-4)=2

=>9x^2-12x+4-9(x^2-16)=2

=>9x^2-12x+4-9x^2+144=2

=>-12x+148=2

=>-12x=-146

=>x=146/12=73/6

d: x^3-6x^2+12x-8=0

=>(x-2)^3=0

=>x-2=0

=>x=2

e: x^3-9x^2+27x-27=0

=>(x-3)^3=0

=>x-3=0

=>x=3

a) \(-25+49x^2=0\)

\(\Leftrightarrow49x^2-25=0\)

\(\Leftrightarrow\left(7x\right)^2-5^2=0\)

\(\Leftrightarrow\left(7x-5\right)\left(7x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}7x-5=0\\7x+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}7x=5\\7x=-5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{7}\\x=-\dfrac{5}{7}\end{matrix}\right.\)

b) \(16x^2-25\left(x-2\right)^2=0\)

\(\Leftrightarrow\left(4x\right)^2-\left[5\left(x-2\right)\right]^2=0\)

\(\Leftrightarrow\left(4x-5x+10\right)\left(4x+5x-10\right)=0\)

\(\Leftrightarrow\left(10-x\right)\left(9x-10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}10-x=0\\9x=10\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=10\\x=\dfrac{10}{9}\end{matrix}\right.\)

c) \(\left(3x-2\right)^2-9\left(x+4\right)\left(x+4\right)=2\)

\(\Leftrightarrow9x^2-12x+4-9\left(x^2+8x+16\right)=2\)

\(\Leftrightarrow9x^2-12x+4-9x^2-72x-144=2\)

\(\Leftrightarrow-84x-140=2\)

\(\Leftrightarrow-84x=142\)

\(\Leftrightarrow x=-\dfrac{142}{84}\)

\(\Leftrightarrow x=-\dfrac{71}{42}\)

d) \(x^3-6x^2+12x-8=0\)

\(\Leftrightarrow x^3-3\cdot2\cdot x^2+3\cdot2^2\cdot x-2^3=0\)

\(\Leftrightarrow\left(x-2\right)^3=0\)

\(\Leftrightarrow x-2=0\)

\(\Leftrightarrow x=2\)

e) \(-27+27x-9x^2+x^3=0\)

\(\Leftrightarrow x^3-9x^2+27x-27=0\)

\(\Leftrightarrow\left(x-3\right)^3=0\)

\(\Leftrightarrow x-3=0\)

\(\Leftrightarrow x=3\)

Mình sửa lại câu c một chút nha:

c: (3x-2)^2-9(x+4)(x+4)=2

=>(3x-2)^2-9(x+4)^2=2

=>(3x-2)^2-(3x+12)^2=2

=>(3x-2-3x-12)(3x-2+3x+12)=2

=>-14*(6x+10)=2

=>6x+10=-1/7

=>6x=-71/7

=>x=-71/42