x = 101 

x = 90

x-91=11

x=11+91

x=102

x-6=93+3

x=96+6

x=102

k nha

a) 135 + x -135 = 0 => (135 - 135) + x = 0 => x = 0.

b) -562 - x + 562 = 0 => (-562 + 562) - x = 0 => - x = 0 => x = 0.

c) 61 + x - 23 = 91=> (61 - 23) + x = 91=>38 + x = 91=> x = 53.

d) 11 + 53 - x = 97 => 64 - x = 97 => x = 64 - 97 => x = - 33

Vì x⋮39, x⋮65, x⋮91 nên x ϵ B(39,65,91)                                                     39=3.13     65=5.13     91=7.13                                                                BCNN(39,65,91)=13^3=2197  BC(39,65,91)=B(2197)=(0,2197,4394...)   mà 400<x<2600 nên xϵ 2197                                                                vì không dùng được dấu ngoặc nhọn nên dùng ngoặc tròn

\(\frac{x-11}{95}+\frac{x-13}{93}=\frac{x-15}{91}+\frac{x-17}{89}\) => \(\frac{x-11}{95}-1+\frac{x-13}{93}-1=\frac{x-15}{91}-1+\frac{x-17}{89}-1\)

=>\(\frac{x-106}{95}+\frac{x-106}{93}=\frac{x-106}{91}+\frac{x-106}{89}\)

=>\(\left(\frac{1}{95}+\frac{1}{93}\right)\left(x-106\right)-\left(\frac{1}{91}+\frac{1}{89}\right)\left(x-106\right)=0\)

<=>\(\left[\left(\frac{1}{95}+\frac{1}{93}\right)-\left(\frac{1}{91}+\frac{1}{89}\right)\right]\left(x-106\right)=0\).Vì\(\frac{1}{95}< \frac{1}{91};\frac{1}{93}< \frac{1}{89}\) nên\(\frac{1}{95}+\frac{1}{93}< \frac{1}{91}+\frac{1}{89}\)

=>\(\left(\frac{1}{95}+\frac{1}{93}\right)-\left(\frac{1}{91}+\frac{1}{89}\right)< 0\) hay khác 0.Vậy x - 106 = 0, tìm được x = 106

a: Ta có: \(\frac{11}{8}+\frac{13}{6}=\frac{85}{2x}\)

=>\(\frac{85}{2x}=\frac{33}{24}+\frac{52}{24}=\frac{85}{24}\)

=>2x=24

=>x=12

b: \(\frac{3x}{182}=\frac{-6}{14}\cdot\frac{35}{91}\)

=>\(\frac{3x}{182}=\frac{-3}{7}\cdot\frac{5}{13}=\frac{-15}{91}\)

=>\(3x=182\cdot\frac{-15}{91}=-30\)

=>x=-10

1: ĐKXĐ: x>=0; x<>9

\(P=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}+1}{\sqrt{x}-3}+\frac{3-11\sqrt{x}}{9-x}\)

\(=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)+11\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(=\frac{2x-6\sqrt{x}+x+4\sqrt{x}+3+11\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{3x+9\sqrt{x}}{\left.\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)\right.}\)

\(=\frac{3\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{3\sqrt{x}}{\sqrt{x}-3}\)

2: Sửa đề: \(x=7+4\sqrt3\)

\(x=7+4\sqrt3\)

\(=4+2\cdot2\cdot\sqrt3+3=\left(2+\sqrt3\right)^2\)

\(P=\frac{3\sqrt{x}}{\sqrt{x}-3}=\frac{3\left(2+\sqrt3\right)}{2+\sqrt3-3}=\frac{3\left(2+\sqrt3\right)}{\sqrt3-1}\)

\(=\frac{3\left(2+\sqrt3\right)\left(\sqrt3+1\right)}{\left(\sqrt3-1\right)\left(\sqrt3+1\right)}=\frac{3\left(2\sqrt3+2+3+\sqrt3\right)}{3-1}=\frac{3\left(3\sqrt3+5\right)}{2}=\frac{9\sqrt3+15}{2}\)

3: P<1

=>P-1<0

=>\(\frac{3\sqrt{x}}{\sqrt{x}-3}-1<0\)

=>\(\frac{3\sqrt{x}-\sqrt{x}+3}{\sqrt{x}-3}<0\)

=>\(\frac{2\sqrt{x}+3}{\sqrt{x}-3}\) <0

=>\(\sqrt{x}-3<0\)

=>\(\sqrt{x}<3\)

=>0<=x<9

a, \(x\) - \(\dfrac{5}{7}\) = \(\dfrac{1}{9}\) 

    \(x\)        = \(\dfrac{1}{9}\) + \(\dfrac{5}{7}\)

    \(x\)        = \(\dfrac{52}{63}\)

b, \(\dfrac{2x}{5}\) = \(\dfrac{6}{2x+1}\)

     2\(x\).(2\(x\) + 1) = 30

     4\(x^2\)+ 2\(x\)  - 30 = 0

  4\(x^2\) + 12\(x\) - 10\(x\) - 30 = 0

   (4\(x^2\) + 12\(x\)) - (10\(x\) + 30) =0

   4\(x\).(\(x\) + 3) - 10.(\(x\) +3) = 0

        2 (\(x\) + 3).(2\(x\) -  5) = 0

            \(\left[{}\begin{matrix}x+3=0\\2x-5=0\end{matrix}\right.\)

             \(\left[{}\begin{matrix}x=-3\\x=\dfrac{5}{2}\end{matrix}\right.\)

Vậy \(x\) \(\in\) {-3; \(\dfrac{5}{2}\)}

ai có câu trả lời nhanh nhất sẽ được tick nha !

Bài 2.Tínhnhẩm: