Bài 1: 

a) Ta có: \(\dfrac{17}{6}-x\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)

\(\Leftrightarrow\dfrac{17}{6}-x^2+\dfrac{7}{6}x-\dfrac{7}{4}=0\)

\(\Leftrightarrow-x^2+\dfrac{7}{6}x+\dfrac{13}{12}=0\)

\(\Leftrightarrow-12x^2+14x+13=0\)

\(\Delta=14^2-4\cdot\left(-12\right)\cdot13=196+624=820\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{14-2\sqrt{205}}{-24}=\dfrac{-7+\sqrt{205}}{12}\\x_2=\dfrac{14+2\sqrt{2015}}{-24}=\dfrac{-7-\sqrt{205}}{12}\end{matrix}\right.\)

b) Ta có: \(\dfrac{3}{35}-\left(\dfrac{3}{5}-x\right)=\dfrac{2}{7}\)

\(\Leftrightarrow\dfrac{3}{5}-x=\dfrac{3}{35}-\dfrac{10}{35}=\dfrac{-7}{35}=\dfrac{-1}{5}\)

hay \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)

ai giúp mik vs

Bài 3: 

a) Ta có: \(2x-3=x+\dfrac{1}{2}\)

\(\Leftrightarrow2x-x=\dfrac{1}{2}+3\)

\(\Leftrightarrow x=\dfrac{7}{2}\)

b) Ta có: \(4x-\left(x+\dfrac{1}{2}\right)=2x-\left(\dfrac{1}{2}-5\right)\)

\(\Leftrightarrow3x-\dfrac{1}{2}-2x+\dfrac{1}{2}-5=0\)

\(\Leftrightarrow x=5\)

a) 78 - x = 90

x = 78 - 90

x = -12

b) 15 + 2x = 5¹⁰ : 5⁸

15 + 2x = 5²

15 + 2x = 25

2x = 25 - 15

2x = 10

x = 10 : 2

x = 5

c) 48 : x + 17 = -33

48 : x + 17 = -33 - 17

48 : x = -50

x = 48 : (-50)

x = -24/25

d) Do x ⋮ 15 và x ⋮ 20

⇒ x ∈ BC(15; 20)

Ta có:

15 = 3.5

20 = 2².5

⇒ BCNN(15; 20) = 2².3.5 = 60

⇒ x ∈ BC(15; 20) = {0; 60; 120; ...}

Mà 50 < x < 70

⇒ x = 60

Bài 1: Tính

a) Ta có: \(\left(-25\right)\cdot68+\left(-34\right)\cdot\left(-250\right)\)

\(=-25\cdot68+\left(-340\right)\cdot\left(-25\right)\)

\(=-25\cdot\left(68-340\right)\)

\(=-25\cdot\left(-272\right)\)

\(=6800\)

b) Ta có: \(1999+\left(-2000\right)+2001+\left(-2002\right)\)

\(=1999-2000+2001-2002\)

\(=-1-1=-2\)

c) Ta có: \(515+\left[72+\left(-515\right)+\left(-32\right)\right]\)

\(=515+72-515-32\)

\(=40\)

d) Ta có: \(\left(2736-75\right)-2736+175\)

\(=2736-75-2736+175\)

\(=100\)

e) Ta có: \(-2020-\left(157-2020\right)-\left(-257\right)\)

\(=-2020-157+2020+257\)

\(=100\)

Bài 2: Tìm x

a) Ta có: \(x-\left|-2\right|=\left|-18\right|\)

\(\Leftrightarrow x-2=18\)

hay x=20

Vậy: x=20

b) Ta có: \(2x-\left|+14\right|=\left|-14\right|\)

\(\Leftrightarrow2x-14=14\)

\(\Leftrightarrow2x=28\)

hay x=14

Vậy: x=14

c) Ta có: \(\left|x+4\right|+5=20-\left(-12-7\right)\)

\(\Leftrightarrow\left|x+4\right|+5=20+12+7\)

\(\Leftrightarrow\left|x+4\right|=39-5=34\)

\(\Leftrightarrow\left[{}\begin{matrix}x+4=34\\x+4=-34\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=30\\x=-38\end{matrix}\right.\)

Vậy: x∈{30;-38}

d) Ta có: \(15-\left|2-x\right|=\left(-2\right)^2\)

\(\Leftrightarrow15-\left|2-x\right|=4\)

\(\Leftrightarrow\left|2-x\right|=11\)

\(\Leftrightarrow\left[{}\begin{matrix}2-x=11\\2-x=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=13\end{matrix}\right.\)

Vậy: x∈{-9;13}

e) Ta có: \(\left|15-x\right|+\left|-25\right|=\left|-55\right|\)

\(\Leftrightarrow\left|15-x\right|+25=55\)

\(\Leftrightarrow\left|15-x\right|=30\)

\(\Leftrightarrow\left[{}\begin{matrix}15-x=30\\15-x=-30\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-15\\x=45\end{matrix}\right.\)

Vậy: x∈{-15;45}

g) Ta có: \(\left|17-\left(-4\right)\right|+\left|-24-\left(-5\right)\right|=\left|-x+3\right|\)

\(\Leftrightarrow\left|17+4\right|+\left|-24+5\right|=\left|3-x\right|\)

\(\Leftrightarrow\left|3-x\right|=40\)

\(\Leftrightarrow\left[{}\begin{matrix}3-x=40\\3-x=-40\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-37\\x=43\end{matrix}\right.\)

Vậy: x∈{-37;43}

a: =>3^x=3^4*3=3^5

=>x=5

b: =>\(2^{x+1}=2^5\)

=>x+1=5

=>x=4

c: \(\Leftrightarrow3^{x+2-3}=3\)

=>x-1=1

=>x=2

d: \(\Leftrightarrow x^2=\dfrac{32}{2}=16\)

=>x=4 hoặc x=-4

e: (2x-1)^4=81

=>2x-1=3 hoặc 2x-1=-3

=>2x=4 hoặc 2x=-2

=>x=-1 hoặc x=2

f: (2x-6)^4=0

=>2x-6=0

=>x-3=0

=>x=3

a) \(3^x=81\cdot3\)

\(\Rightarrow3^x=3^4\cdot3\)

\(\Rightarrow3^x=3^5\)

\(\Rightarrow x=5\)

b) \(2^{x+1}=32\)

\(\Rightarrow2^{x+1}=2^5\)

\(\Rightarrow x+1=5\)

\(\Rightarrow x=4\)

c) \(3^{x+2}:27=3\)

\(\Rightarrow3^{x+2}:3^3=3\)

\(\Rightarrow3^{x+2-3}=3\)

\(\Rightarrow3^{x-1}=3\)

\(\Rightarrow x-1=1\)

\(\Rightarrow x=2\)

d) \(2x^2=32\)

\(\Rightarrow x^2=16\)

\(\Rightarrow x^2=4^2\)

\(\Rightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)

e) \(\left(2x-1\right)^4=81\)

\(\Rightarrow\left(2x-1\right)^4=3^4\)

\(\Rightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=4\\2x=-2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)

f)  \(\left(2x-6\right)^4=0\)

\(\Rightarrow2x-6=0\)

\(\Rightarrow2x=6\)

\(\Rightarrow x=6:2\)

\(\Rightarrow x=3\)

\(a,3^x=81\cdot3\\ \Leftrightarrow3^x=3^4\cdot3\\ \Leftrightarrow3^x=3^5\\ \Leftrightarrow x=5\\ d,2^{x+1}=32\\ \Leftrightarrow x+1=5\\ \Leftrightarrow x=4\\ c,3^{x+2}:27=3\\ \Leftrightarrow3^{x+2}:3^3=3\\ \Leftrightarrow3^{x-1}=3\\ \Leftrightarrow x-1=1\\ \Leftrightarrow x=2\\ d,2x^2=32\\ \Leftrightarrow x^2=16\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\\ e,\left(2x-1\right)^4=81\\ \Leftrightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\\ f,\left(2x-6\right)^4=0\\ \Leftrightarrow2x-6=0\\ \Leftrightarrow x=3\)

a: Sửa đề: \(\dfrac{12}{-6}=\dfrac{x}{5}=\dfrac{-y}{3}=\dfrac{2}{-z}=\dfrac{-t}{-9}\)

=>\(\dfrac{x}{5}=\dfrac{y}{-3}=\dfrac{-2}{z}=\dfrac{t}{9}=-2\)

=>\(x=-2\cdot5=-10;y=-2\cdot\left(-3\right)=6;z=\dfrac{-2}{-2}=1;t=9\cdot\left(-2\right)=-18\)

b: \(\dfrac{-24}{-6}=\dfrac{x}{3}=\dfrac{4}{y^2}=\dfrac{z^3}{-2}\)

=>\(\dfrac{x}{3}=\dfrac{4}{y^2}=\dfrac{z^3}{-2}=4\)

=>\(\left\{{}\begin{matrix}x=4\cdot3=12\\y^2=\dfrac{4}{4}=1\\z^3=-2\cdot4=-8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=12\\y\in\left\{1;-1\right\}\\z=-2\end{matrix}\right.\)

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 \(x\in\left\{2;3\right\}\)

a: =>2x-x=-5/2-1/3

=>x=-17/6

b: =>4(x-2)²=36

=>(x-2)²=9

=>x-2=3 hoặc x-2=-3

hay x=5 hoặc x=-1

c: =>2x+1/2=5/6

=>2x=1/3

hay x=1/6

a: =>2x-x=-5/2-1/3

=>x=-17/6

b: =>4(x-2)²=36

=>(x-2)²=9

=>x-2=3 hoặc x-2=-3

hay x=5 hoặc x=-1

c: =>2x+1/2=5/6

=>2x=1/3

hay x=1/6