`#040911`

`a)`

`3 1/3 x + 16 3/4 = -13,25`

`=> 3 1/3 x = -13,25 - 16 3/4`

`=> 3 1/3 x = -30`

`=> x = -30 \div 3 1/3`

`=> x =-9`

Vậy, `x = -9`

`b)`

`3 2/7*x - 1/8 = 2 3/4`

`=> 3 2/7x = 2 3/4 + 1/8`

`=> 3 2/7x = 23/8`

`=> x = 23/8 \div 3 2/7`

`=> x = 7/8`

Vậy, `x = 7/8`

`c)`

`x \div 4 1/3 = -2,5`

`=> x = -2,5 * 4 1/3`

`=> x = -65/6`

Vậy, `x = -65/6`

`d)`

`( (3x)/7 + 1) \div (-4) = (-1)/28`

`=> (3x)/7 +1 = (-1)/28 * (-4)`

`=> (3x)/7 + 1 = 1/7`

`=> (3x)/7 = 1/7 - 1`

`=> (3x)/7 = -6/7`

`=> 3x = -6`

`=> x = -6 \div 3`

`=> x = -2`

Vậy, `x = -2.`

a

=>10/3 . x + 16 + 3/4 = -13,25

=>10/3 x + 3/4 = -29,25

=>10/3 x = -30

=>x=-30 : 10/3

=>x=-30 . 3/10

=>x=-9

b.

=>23/7 x - 1/8 = = 11/4

=>23/7 x = 11/4 + 1/8

=>23/7 x= 22/8 + 1/8

=>23/7 x= 23/8

=>x=23/8 : 23/7

=>x=23/8 . 7/23

=>x=7/8

c.

=>x : 13/3 =-5/2

=>x=-5/2 . 13/3

=>x=-65/6

d.

=>3x/7 +1 = (-1/28) . (-4)

=>3x/7 + 1 = 1/7

=>3x/7 = -6/7

=>3x=-6

=>x=-2

a) \(3\dfrac{1}{3}x+16\dfrac{3}{4}=-13,25\)

\(\Rightarrow\dfrac{10}{3}x+\dfrac{67}{4}=-\dfrac{53}{4}\)

\(\Rightarrow\dfrac{10}{3}x=-30\)

\(\Rightarrow x=-30:\dfrac{10}{3}\)

\(\Rightarrow x=-9\)

b) \(3\dfrac{2}{7}x-\dfrac{1}{8}=2\dfrac{3}{4}\)

\(\Rightarrow\dfrac{23}{7}x-\dfrac{1}{8}=\dfrac{11}{4}\)

\(\Rightarrow\dfrac{23}{7}x=\dfrac{11}{4}+\dfrac{1}{8}\)

\(\Rightarrow\dfrac{23}{7}x=\dfrac{23}{8}\)

\(\Rightarrow x=\dfrac{23}{8}:\dfrac{23}{7}\)

\(\Rightarrow x=\dfrac{7}{8}\)

c) \(x:4\dfrac{1}{3}=-2,5\)

\(\Rightarrow x:\dfrac{13}{3}=-\dfrac{5}{2}\)

\(\Rightarrow x=-\dfrac{5}{2}\cdot\dfrac{13}{3}\)

\(\Rightarrow x=-\dfrac{65}{6}\)

d) \(\left(\dfrac{3x}{7}+1\right):\left(-4\right)=\dfrac{-1}{28}\)

\(\Rightarrow\dfrac{3x}{7}+1=\dfrac{-1}{28}\cdot-4\)

\(\Rightarrow\dfrac{3x}{7}+1=\dfrac{1}{7}\)

\(\Rightarrow\dfrac{3x}{7}=-\dfrac{6}{7}\)

\(\Rightarrow x=-\dfrac{6}{7}:\dfrac{3}{7}\)

\(\Rightarrow x=-2\)

a: Ta có: \(\left(2x+1\right)^2-4\left(x+2\right)^2=9\)

\(\Leftrightarrow4x^2+4x+1-4x^2-16x-16=9\)

\(\Leftrightarrow-12x=24\)

hay x=-2

b: Ta có: \(\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)=1\)

\(\Leftrightarrow x^2+6x+9-x^2-4x+32=1\)

\(\Leftrightarrow2x=-40\)

hay x=-20

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

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

\(\Leftrightarrow x=5\)

Lời giải:
a.

$0< x< \frac{1}{4}+\frac{4}{5}$

$\Rightarrow 0< x< \frac{21}{20}$ hay $0< x< 1,05$

$\Rightarrow x=1$

b.

$\frac{4}{7}+\frac{3}{7}< x< \frac{5}{3}+\frac{2}{3}$
$\Rightarrow 1< x< \frac{7}{3}$
$\Rightarrow x=2$

cha loi di

Lời giải:

a. ĐKXĐ: $x\geq -9$

PT $\Leftrightarrow x+9=7^2=49$

$\Leftrightarrow x=40$ (tm)

b. ĐKXĐ: $x\geq \frac{-3}{2}$

PT $\Leftrightarrow 4\sqrt{2x+3}-\sqrt{4(2x+3)}+\frac{1}{3}\sqrt{9(2x+3)}=15$

$\Leftrightarrow 4\sqrt{2x+3}-2\sqrt{2x+3}+\sqrt{2x+3}=15$

$\Leftrgihtarrow 3\sqrt{2x+3}=15$

$\Leftrightarrow \sqrt{2x+3}=5$

$\Leftrightarrow 2x+3=25$

$\Leftrightarrow x=11$ (tm)

c.

PT \(\Leftrightarrow \left\{\begin{matrix} 2x+1\geq 0\\ x^2-6x+9=(2x+1)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq \frac{-1}{2}\\ 3x^2+10x-8=0\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\geq \frac{-1}{2}\\ (3x-2)(x+4)=0\end{matrix}\right.\)

\(\Leftrightarrow x=\frac{2}{3}\)

d. ĐKXĐ: $x\geq 1$

PT \(\Leftrightarrow \sqrt{(x-1)+4\sqrt{x-1}+4}-\sqrt{(x-1)+6\sqrt{x-1}+9}=9\)

\(\Leftrightarrow \sqrt{(\sqrt{x-1}+2)^2}-\sqrt{(\sqrt{x-1}+3)^2}=9\)

\(\Leftrightarrow \sqrt{x-1}+2-(\sqrt{x-1}+3)=9\)

\(\Leftrightarrow -1=9\) (vô lý)

Vậy pt vô nghiệm.

a) \(\sqrt{x+9}=7\left(x\ge-9\right)\Rightarrow x+9=49\Rightarrow x=40\)

b) \(4\sqrt{2x+3}-\sqrt{8x+12}+\dfrac{1}{3}\sqrt{18x+27}=15\left(x\ge-\dfrac{3}{2}\right)\)

\(\Rightarrow4\sqrt{2x+3}-\sqrt{4\left(2x+3\right)}+\dfrac{1}{3}\sqrt{9\left(2x+3\right)}=15\)

\(\Rightarrow4\sqrt{2x+3}-2\sqrt{2x+3}+\sqrt{2x+3}=15\)

\(\Rightarrow3\sqrt{2x+3}=15\Rightarrow\sqrt{2x+3}=5\Rightarrow2x+3=25\Rightarrow x=11\)

c) \(\sqrt{x^2-6x+9}=2x+1\)

Vì \(VT\ge0\Rightarrow VP\ge0\Rightarrow x\ge-\dfrac{1}{2}\)

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

\(\Rightarrow\left[{}\begin{matrix}x=-4\left(l\right)\\x=\dfrac{2}{3}\end{matrix}\right.\)

d) \(\sqrt{x+3+4\sqrt{x-1}}-\sqrt{x+8+6\sqrt{x-1}}=9\left(x\ge1\right)\)

\(\Rightarrow\sqrt{x-1+4\sqrt{x-1}+4}-\sqrt{x-1+6\sqrt{x-1}+9}=9\)

\(\Rightarrow\sqrt{\left(\sqrt{x-1}+2\right)^2}-\sqrt{\left(\sqrt{x-1}+3\right)^2}=9\)

\(\Rightarrow\left|\sqrt{x-1}+2\right|-\left|\sqrt{x-1}+3\right|=9\)

\(\Rightarrow\sqrt{x-1}+2-\sqrt{x-1}-3=9\Rightarrow-1=9\) (vô lý)

a) \(\sqrt{\left(2x-3\right)^2}=7\)

\(\Leftrightarrow\left|2x-3\right|=7\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}2x=10\\2x=-4\end{matrix}\right.\)

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

b) \(\sqrt{64x+128}-\sqrt{25x+50}+\sqrt{4x+8}=20\left(đk:x\ge-2\right)\)

\(\Leftrightarrow8\sqrt{x+2}-5\sqrt{x+2}+2\sqrt{x+2}=20\)

\(\Leftrightarrow5\sqrt{x+2}=20\)

\(\Leftrightarrow\sqrt{x+2}=4\Leftrightarrow x+2=16\Leftrightarrow x=14\left(tm\right)\)

c) \(\sqrt{x^2-9}-3\sqrt{x-3}=0\left(đk:x\ge3\right)\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\\sqrt{x+3}=3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x+3=9\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=6\left(tm\right)\end{matrix}\right.\)

a. \(\sqrt{\left(2x-3\right)^2}=7\)

<=> \(\left|2x-3\right|=7\)

<=> \(\left[{}\begin{matrix}2x-3=7\left(x\ge\dfrac{3}{2}\right)\\-2x+3=7\left(x< \dfrac{3}{2}\right)\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}2x=10\\-2x=4\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=5\left(TM\right)\\x=-2\left(TM\right)\end{matrix}\right.\)

b. \(\sqrt{64x+128}-\sqrt{25x+50}+\sqrt{4x+8}=20\)  ĐK: \(x\ge-2\)

<=> \(\sqrt{64\left(x+2\right)}-\sqrt{25\left(x+2\right)}+\sqrt{4\left(x+2\right)}-20=0\)

<=> \(8\sqrt{x+2}-5\sqrt{x+2}+2\sqrt{x+2}-20=0\)

<=> \(\sqrt{x+2}.\left(8-5+2\right)-20=0\)

<=> \(5\sqrt{x+2}=20\)

<=> \(\sqrt{x+2}=4\)

<=> \(\left(\sqrt{x+2}\right)^2=4^2\)

<=> \(\left|x+2\right|=16\)

<=> \(\left[{}\begin{matrix}x+2=16\left(x\ge-2\right)\\x+2=-16\left(x< -2\right)\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=14\left(TM\right)\\x=-18\left(TM\right)\end{matrix}\right.\)

c. \(\sqrt{x^2-9}-3\sqrt{x-3}=0\)             ĐK: \(x\ge3\)

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

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

<=> \(\left(\sqrt{x+3}-3\right).\sqrt{x-3}=0\)

<=> \(\left[{}\begin{matrix}\sqrt{x+3}-3=0\\\sqrt{x-3}=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=6\\x=3\end{matrix}\right.\)

a) I2x-3I=7

        2x-3=7     =>x=5

        2x-3=-7     =>x=-2

b) \(8\sqrt{3x}-5\sqrt{3x}+2\sqrt{3x}=20\)

   5\(\sqrt{3x}=20\)

   3x=16  =>x=16/3

c)  vì câu c dài nên mình chỉ cho đáp án thôi là 0,3,6

vì \(\sqrt{ }\) của 1 số luôn dương nên 3,6 thỏa mãn 

1:

a: \(=\dfrac{-4}{7}+\dfrac{4}{7}+\dfrac{3}{7}-\dfrac{23}{34}-\dfrac{4}{5}=\dfrac{3}{7}-\dfrac{23}{34}-\dfrac{4}{5}=-\dfrac{1247}{1190}\)

b:

Sửa đề: \(\dfrac{-5}{13}+\dfrac{4}{19}+\dfrac{-8}{13}+\dfrac{15}{19}+\dfrac{45}{6}\) 

\(=\dfrac{-5}{13}-\dfrac{8}{13}+\dfrac{4}{19}+\dfrac{15}{19}+\dfrac{45}{6}=\dfrac{9}{2}\)

a) `4\sqrt(2x-1)>8`

`<=>\sqrt(2x-1)>2`

`<=>2x-1>4`

`<=>x>5/2`

b) `2\sqrtx-1>3`

`<=>2\sqrtx>4`

`<=>\sqrtx>2`

`<=>x>4`

a) Ta có: \(4\sqrt{2x-1}>8\)

\(\Leftrightarrow2x-1>4\)

\(\Leftrightarrow2x>5\)

hay \(x>\dfrac{5}{2}\)

b) Ta có: \(2\sqrt{x}-1>3\)

\(\Leftrightarrow\sqrt{x}>2\)

hay x>4

a: \(-4x\left(x-5\right)-2x\left(8-2x\right)=-3\)

=>\(-4x^2+20x-16x+4x^2=-3\)

=>4x=-3

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

b: \(-7\left(x+9\right)-3\left(5-x\right)=2\)

=>\(-7x-63-15+3x=2\)

=>\(-4x-78=2\)

=>\(-4x=78+2=80\)

=>\(x=\dfrac{80}{-4}=-20\)