Bài 1.

a, (x-2)-15=65

x-2=65+15

x-2=80

x=80+2

x=2

b, 115-2\(\times\)(x-3)=35

2\(\times\)(x-3)=115-35

2\(\times\)(x-3)=70

x-3=70:2

x-3=35

x=35+5

x=38

c, 35+2\(\times\)(x-3)=65

2\(\times\)(x-3)=65-35

2\(\times\)(x-3)=30

x-3=30:2

x-3=15

x=15+3

x=18

3\(\times\)(x-5)-16=11

3\(\times\)(x-5)=11+16

3\(\times\)(x-5)=27

x-5=27:3

x-5=9

x=9+5

x=14

Bài 2:

a, \(2^x-1=31\)

\(2^x=31-1\)

\(2^x=30\)

\(\Rightarrow\)Không có x thoả mãn điều kiện \(2^x=30\)

b, \(x^3-1=26\)

\(x^3=26+1\)

\(x^3=27\)

\(\Rightarrow x=3\) vì \(3^3=27\)

c, \(6^x-1+1=37\)

\(6^x-1=37-1\)

\(6^x-1=36\)

\(6^x=36+1\)

\(6^x=37\)

\(\Rightarrow\) Không có x thoả mãn điều kiện \(6^x=37\)

d, (x+2)\(^3\)-15\(^0\)=215

\(\left(x+2\right)^3-1=215\)

\(\left(x+2\right)^3=215+1\)

\(\left(x+2\right)^3=216\)

\(\left(x+2\right)^3=6^3\)

\(x+2=6\)

\(x=6-2\)

\(x=4\)

e, \(2\times\left(x-9\right)^2=2\)

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

\(\left(x-9\right)^2=1\)

\(\Rightarrow x-9=1\) vì \(1^2=1\)

x=1+9

x=10

g, \(3\times\left(x-5\right)^3=51\)

\(\left(x-5\right)^3=51:3\)

\(\left(x-5\right)^3=17\)

\(\Rightarrow\) Không có x thoả mãn điều kiện \(\left(x-5\right)^3=17\)

Nếu đúng thì tick cho mk nhé

Bài 1:

a)\(\left(x-2\right)-15=65\)

\(x-2=65+15\)

\(x-2=80\)

\(x=80+2\)

\(x=82\)

b)\(115-2\left(x-3\right)=35\)

\(2\left(x-3\right)=115-35\)

\(2\left(x-3\right)=80\)

\(x-3=80:2\)

\(x-3=40\)

\(x=40+3\)

c) \(35+2\left(x-3\right)=65\)

\(2\left(x-3\right)=65-35=30\)

\(x-3=30:2=15\)

\(x=15+3=18\)

d) \(3\left(x-5\right)-16=11\)

\(3\left(x-5\right)=11+16=27\)

\(x-5=27:3=9\)

\(x=9+5=14\)

Bài 2:

a) \(2^x-1=31\)

\(2^x=31+1=32\)

Vì \(2^5=32\Rightarrow x=5\)

b) \(x^3-1=26\)

\(x^3=26+1=27\)

Vì \(3^3=27\Rightarrow x=3\)

c)\(6^{x-1}+1=37\)

\(6^{x-1}=37-1=36\)

Vì \(6^6=36\Rightarrow x-1=6\Rightarrow x=6+1=7\)

d)\(\left(x+2\right)^3-15^0=215\)

\(\left(x+2\right)^3-1=215\)

\(\left(x+2\right)^3=215+1=216\)

Vì \(6^3=216\Rightarrow x+2=6\Rightarrow x=6-2=4\)

e)\(2\left(x-9\right)^2=2\)

\(\left(x-9\right)^2=2:2=1\)

Vì \(1^2=1\Rightarrow x-9=1\Rightarrow x=1+9=10\)

g) \(3\left(x-5\right)^3=51\)

\(\left(x-5\right)^3=51:3=17\)

Bài 1:

a)

\(\left(x-2\right)-15=65\)

\(x-2\) \(=65+15\)

\(x-2\) \(=\) \(80\)

\(x\) \(=80+2\)

\(x\) \(=82\)

b)

\(115-2\cdot\left(x-3\right)=35\)

\(2\cdot\left(x-3\right)=115-35\)

\(2\cdot\left(x-3\right)=80\)

\(x-3=80:2\)

\(x-3=40\)

\(x\) \(=40+3\)

\(x\) \(=43\)

c)

\(35+2\cdot\left(x-3\right)=65\)

\(2\cdot\left(x-3\right)=65-35\)

\(2\cdot\left(x-3\right)=30\)

\(x-3=30:2\)

\(x-3=15\)

\(x\) \(=15+3\)

\(x\) \(=18\)

d)

\(3\cdot\left(x-5\right)-16=11\)

\(3\cdot\left(x-5\right)\) \(=11+16\)

\(3\cdot\left(x-5\right)\) \(=27\)

\(x-5\) \(=27:3\)

\(x-5\) \(=9\)

\(x\) \(=9+5\)

\(x\) \(=14\)

Bài 2

a)

\(2^x-1=31\)

\(2^x\) \(=31+1\)

\(2^x\) \(=32\)

\(2^x\) \(=2^5\)

⇒ \(x=5\)

b)

\(x^3-1=26\)

\(x^3\) \(=26+1\)

\(x^3\) \(=27\)

\(x^3\) \(=3^3\)

⇒ \(x=3\)

c)

\(6^{x-1}+1=37\)

\(6^{x-1}\) \(=37-1\)

\(6^{x-1}\) \(=36\)

\(6^{x-1}\) \(=6^2\)

➞ \(x-1=2\)

\(x\) \(=2+1\)

\(x\) \(=3\)

d)

\(\left(x+2\right)^3-15^0=215\)

\(\left(x+2\right)^3-1=215\)

\(\left(x+2\right)^3=215+1\)

\(\left(x+2\right)^3=216\)

\(\left(x+2\right)^3=6^3\)

➞ \(x+2=6\)

\(x=6-2\)

\(x=4\)

e)

\(2\cdot\left(x-9\right)^2=2\)

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

\(\left(x-9\right)^2=1\)

\(\left(x-9\right)^2=1^2\)

➞ \(x-9=1\)

\(x=9+1\)

\(x=10\)

g)

\(3\cdot\left(x-5\right)^3=51\)

Câu hỏi này sai nhé bạn

Bài 1:

A.\(\left(\sqrt{x}+2\right)\) = -1 (ĐK: \(x\ge0\)

\(\Leftrightarrow\dfrac{1}{x-4}\left(\sqrt{x}+2\right)=-1\)

\(\Leftrightarrow\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=-1\)

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

\(\Leftrightarrow\sqrt{x}-2=-1\)

\(\Leftrightarrow\sqrt{x}=1\\ \Leftrightarrow x=1\left(TM\right)\)

Vậy x = 1

Bài 2: ĐK: \(x\ge0\)

Để \(B\in Z\Leftrightarrow\dfrac{3}{\sqrt{x}+2}\in Z\Leftrightarrow\sqrt{x}+2\inƯ\left(3\right)\)\(\Leftrightarrow\sqrt{x}+2\in\left\{\pm1,\pm3\right\}\)\(\Leftrightarrow x\in\left\{1\right\}\)

Bài 3:

a, Ta có: \(x+\sqrt{x}+1=x+2.\dfrac{1}{2}\sqrt{x}+\dfrac{1}{4}-\dfrac{1}{4}+1\\ =\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\)

Ta có: 2 > 0 và \(x+\sqrt{x}+1>0\Rightarrow C>0\) và \(x\ne1\)

b, ĐK: \(x\ge0,x\ne1\)

\(C=\dfrac{2}{x+\sqrt{x}+1}\)

Ta có: \(x+\sqrt{x}+1=\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)

Ta có: \(\sqrt{x}\ge0\forall x\Rightarrow\sqrt{x}+\dfrac{1}{2}\ge\dfrac{1}{2}\forall x\Leftrightarrow\left(\sqrt{x}+\dfrac{1}{2}\right)^2\ge\dfrac{1}{4}\)

\(\Leftrightarrow\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge1\Leftrightarrow\dfrac{2}{\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le2\)

Dấu bằng xảy ra \(\Leftrightarrow\sqrt{x}+\dfrac{1}{2}=\dfrac{1}{2}\\ \Leftrightarrow x=0\left(TM\right)\)

Vậy MaxC = 2 khi x = 0

Còn cái GTNN chưa tính ra được, để sau nha

Bài 4: ĐK: \(x\ge0,x\ne1\)

\(D=\left(\dfrac{2x+1}{\sqrt{x^3-1}}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\dfrac{1+\sqrt{x^3}}{1+\sqrt{x}}-\sqrt{x}\right)\)

\(=\left(\dfrac{2x+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\dfrac{\left(1+\sqrt{x}\right)\left(x-\sqrt{x}+1\right)}{1+\sqrt{x}}-\sqrt{x}\right)\)

\(=\left(\dfrac{2x+1-\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\left(x-\sqrt{x}+1-\sqrt{x}\right)\)

\(=\left(\dfrac{2x+1-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\left(x-2\sqrt{x}+1\right)\)

\(=\left(\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\left(\sqrt{x}-1\right)^2\)

\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)}\)

\(=\sqrt{x}-1\)

\(D=3\Leftrightarrow\sqrt{x}-1=3\Leftrightarrow x=2\left(TM\right)\)

\(D=x-3\sqrt{x}+2\)

\(\Leftrightarrow\sqrt{x}-1=\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)\)

\(\Leftrightarrow\left(\sqrt{x}-1\right)-\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=0\)

\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(1-\sqrt{x}+2\right)=0\)

\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(3-\sqrt{x}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(L\right)\\x=9\left(TM\right)\end{matrix}\right.\)

Bài 5: \(E< -1\Leftrightarrow\dfrac{-3x}{2x+4\sqrt{x}}< -1\)\(\Leftrightarrow\dfrac{-3x}{2x+4\sqrt{x}}+1< 0\Leftrightarrow\dfrac{-3x+2x+4\sqrt{x}}{2x+4\sqrt{x}}< 0\)

\(\Leftrightarrow\dfrac{4\sqrt{x}-x}{2x+4\sqrt{x}}< 0\Leftrightarrow\dfrac{\sqrt{x}\left(4-\sqrt{x}\right)}{2x+4\sqrt{x}}< 0\)

Ta có: \(\sqrt{x}>0\Leftrightarrow x>0\Leftrightarrow2x+4\sqrt{x}>0\) mà \(\dfrac{\sqrt{x}\left(4-\sqrt{x}\right)}{2x+4\sqrt{x}}< 0\)\(\Rightarrow\sqrt{x}\left(4-\sqrt{x}\right)< 0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}< 0\left(L\right)\\4-\sqrt{x}>0\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}>0\\4-\sqrt{x}< 0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x< 16,x\ne0\\\left\{{}\begin{matrix}x>0\\x< 16\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x< 16,x\ne0\\0< x< 16\end{matrix}\right.\)

a) \(\frac{2}{3}+\frac{1}{3}:x=\frac{3}{5}\)

\(\frac{1}{3}:x=\frac{3}{5}-\frac{2}{3}=\frac{9}{15}-\frac{10}{15}=\frac{-1}{15}\)

\(x=\frac{-1}{15}.\frac{1}{3}\)

\(x=\frac{-1}{45}\)

Vậy x = \(\frac{-1}{45}\)

c) \(\left|2x-1\right|+1=4\)

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

2x-1 = 3 ; -3

TH1: 2.x - 1 = 3

2.x = 3 + 1 = 4

x = 4 : 2 = 2

TH2: 2.x - 1 = -3

2.x = -3 + 1 = -2

x = -2 : 2 = -1

Vậy x \(\in\){ 2 ; -1 }

Ngại làm ấn máy ==

Tập xác định của phương trình

Rút gọn thừa số chung

Giải phương trình

Biệt thức

Biệt thức

Nghiệm

Lời giải thu được

Bài 2 :

a, x = \(\dfrac{-3}{-11}\) => x =\(\dfrac{3}{11}\)

=>| x | = \(\dfrac{3}{11}\)

=> x= \(\dfrac{3}{11}\) hoặc x = \(\dfrac{-3}{11}\)

Bài 3 :

a, | 4.(x-1)| =12

=> 4.(x-1)=12 hoặc 4.(x-1)=-12

\(\left[{}\begin{matrix}4.\left(x-1\right)=12\\4.\left(x-1\right)=-12\end{matrix}\right.=>\left[{}\begin{matrix}4x-4=12\\4x-4=-12\end{matrix}\right.=>\left[{}\begin{matrix}4x=16\\4x=-8\end{matrix}\right.=>\left[{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\)

Vậy x = 4 hoặc x = -2

b,|2x+1|-5 =10

|2x+1|=15

=2x+1=15 hoặc 2x+=-15

+) 2x+1=15 = > 2x = 14 = > x =7

+)2x+1=-15 => 2x= -16 => x = -8

Vậy x=7 hoặc x = -8

c,|2,5-x|-1,3=0

|2,5-x|= 1,3

=>2,5 -x = 1,3 hoặc 2,5 - x = -1,3

+)2,5 - x = 1,3 => x = 1,2

+)2,5-x = -1,3 => x=3,8

Vậy x = 1,2 hoặc x = 3,8

d,-|1,4 - x | - 2 = 0

-|1,4-x|=2

=> -1,4+x = 2 hoặc -1,4+x = -2

+) -1,4+x= 2 => x = 3,4

+)-1,4+x= -2 => x = 0,6

Vậy x = 3,4 hoặc x = 0 ,6

e,| x - 2 | = x

=> x -2 = x hoặc x - 2 = -x

+) x- 2 = x => x-x = -2 => 0 = -2 ( vô lí )

+) x -2 = -x => x+x=2 => 2x =2 => x= 1

Vậy x = 1

f, 2.|2x-3| = \(\dfrac{1}{2}\)

=> |2x-3|= \(\dfrac{1}{4}\)

=>2x-3=\(\dfrac{1}{4}\) hoặc 2x-3=\(\dfrac{-1}{4}\)

+) 2x - 3 = \(\dfrac{1}{4}\)=> 2x= \(\dfrac{13}{4}\)=> x = \(\dfrac{13}{8}\)

+) 2x - 3 = \(\dfrac{-1}{4}\)=> 2x=\(\dfrac{11}{4}\)=> x = \(\dfrac{11}{8}\)

Vậy x=\(\dfrac{13}{8}\) hoặc x=\(\dfrac{11}{8}\)

b)

\(\dfrac{1}{2}\left(\dfrac{4}{5}-\dfrac{3}{2}\right)+x=5\left(x-\dfrac{1}{3}\right)\)

=> \(-\dfrac{7}{20}+x=5x-\dfrac{5}{3}\)

=> \(\dfrac{79}{60}=4x\)

=> \(\dfrac{79}{240}=x\)

Vậy \(\dfrac{79}{240}=x\)

c)

\(2\left(x-5\right)-3\left(x+7\right)=14\)

=> \(2x-10-3x-21=14\)

=> \(-x-31=14\)

=> \(-x=45\)

=> x = -45

Vậy x là -45

d)

-8 . |x - 3| = -32

=> |x - 3| = 4

=> \(\left\{{}\begin{matrix}x-3=4\\x-3=-4\end{matrix}\right.\)

=> x = 7 và x = -1

Vậy x \(\in\left\{-1;7\right\}\)

Bài 9 : Tìm x, biết :

a, (x - 2)(x - 3) + (x - 2) - 1 = 0

\(\Leftrightarrow\left(x-2\right)\left(x-3+1\right)-1=0\)

\(\Leftrightarrow\left(x-2\right)^2-1=0\)

\(\Leftrightarrow\left(x-2+1\right)\left(x-2-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

Vậy x ={1; 3}

b, (x + 2)² - 2x(2x + 3) = (x + 1)²

\(\Leftrightarrow\left(x+2\right)^2-\left(x+1\right)^2-2x\left(2x+3\right)=0\)

\(\Leftrightarrow\left(x+2+x+1\right)\left(x+2-x-1\right)-2x\left(2x+3\right)=0\)

\(\Leftrightarrow2x+3-2x\left(2x+3\right)=0\)

\(\Leftrightarrow\left(2x+3\right)\left(1-2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\1-2x=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{3}{2}\\x=\frac{1}{2}\end{matrix}\right.\)

Vậy \(x=\left\{-\frac{3}{2};\frac{1}{2}\right\}\)
c, 6x³ + x² = 2x

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

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

\(\Leftrightarrow x\left(6x^2+4x-3x-2\right)=0\)

\(\Leftrightarrow x\left[2x\left(3x+2\right)-\left(3x+2\right)\right]=0\)

\(\Leftrightarrow x\left(3x+2\right)\left(2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\3x+2=0\\2x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\frac{2}{3}\\x=\frac{1}{2}\end{matrix}\right.\)

Vậy \(x=\left\{0;-\frac{2}{3};\frac{1}{2}\right\}\)

B=\(\dfrac{1}{\left|x-2\right|+3}\)

do \(\left|x-2\right|\ge0\forall x\)

=> \(\left|x-2\right|+3\ge3\)

=> \(\dfrac{1}{\left|x-2\right|+3}\le\dfrac{1}{3}\)

=> B \(\le\dfrac{1}{3}\)

GTLN của B =\(\dfrac{1}{3}\)

khi x-2=0

=> x=2

vậy GTLN của A=\(\dfrac{1}{3}\) khi x=2