\(4\sqrt{x+3}+\sqrt{19-3x}=x^2+2x+9\)

2,

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AH
Akai Haruma
Giáo viên
14 tháng 10 2018

1)

ĐK: \(x\geq 5\)

PT \(\Leftrightarrow \sqrt{4(x-5)}+3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9(x-5)}=6\)

\(\Leftrightarrow \sqrt{4}.\sqrt{x-5}+3\sqrt{\frac{1}{9}}.\sqrt{x-5}-\frac{1}{3}.\sqrt{9}.\sqrt{x-5}=6\)

\(\Leftrightarrow 2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=6\)

\(\Leftrightarrow 2\sqrt{x-5}=6\Rightarrow \sqrt{x-5}=3\Rightarrow x=3^2+5=14\)

AH
Akai Haruma
Giáo viên
14 tháng 10 2018

2)

ĐK: \(x\geq -1\)

\(\sqrt{x+1}+\sqrt{x+6}=5\)

\(\Leftrightarrow (\sqrt{x+1}-2)+(\sqrt{x+6}-3)=0\)

\(\Leftrightarrow \frac{x+1-2^2}{\sqrt{x+1}+2}+\frac{x+6-3^2}{\sqrt{x+6}+3}=0\)

\(\Leftrightarrow \frac{x-3}{\sqrt{x+1}+2}+\frac{x-3}{\sqrt{x+6}+3}=0\)

\(\Leftrightarrow (x-3)\left(\frac{1}{\sqrt{x+1}+2}+\frac{1}{\sqrt{x+6}+3}\right)=0\)

\(\frac{1}{\sqrt{x+1}+2}+\frac{1}{\sqrt{x+6}+3}>0, \forall x\geq -1\) nên $x-3=0$

\(\Rightarrow x=3\) (thỏa mãn)

Vậy .............

25 tháng 10 2020

b, ĐKXĐ: \(x\ge\frac{5}{2}\)

\(pt\Leftrightarrow\sqrt{2x+4-6\sqrt{2x-5}}+\sqrt{2x-4+2\sqrt{2x-5}}=4\)

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

\(\Leftrightarrow\sqrt{2x-5}=3\)

\(\Leftrightarrow x=7\left(tm\right)\)

25 tháng 10 2020

a, ĐKXĐ: \(x\ge5\)

\(pt\Leftrightarrow\sqrt{x-5+4\sqrt{x-5}+4}+\sqrt{x-5+8\sqrt{x-5}+16}=0\)

\(\Leftrightarrow\sqrt{\left(\sqrt{x-5}+2\right)^2}+\sqrt{\left(\sqrt{x-5}+4\right)^2}=0\)

\(\Leftrightarrow2\sqrt{x-5}+6=0\)

\(\Leftrightarrow\sqrt{x-5}=-3\)

Phương trình vô nghiệm

bài 1: giải các hệ phương trình 1)\(\dfrac{1}{x}\)+\(\dfrac{1}{y}\)=\(\dfrac{1}{2}\) x+y=9 2) \(\dfrac{2x+1}{4}-\dfrac{y-2}{3}=\dfrac{1}{12}\) \(\dfrac{x+5}{2}-\dfrac{y+7}{3}=-4\) 3)\(2|x|-y=3\) \(|x|+y=3\) 4)\(2\left(x+y\right)+\sqrt{x+1}=4\) \(\left(x+y\right)-3\sqrt{x+1}=-5\) 5) \(\dfrac{7}{2x+y}+\dfrac{4}{2x-y}=74\) \(\dfrac{3}{2x+y}+\dfrac{2}{2x-y}=32\) 6)\(\dfrac{1}{x}+\dfrac{3}{2y+1}=2\) \(\dfrac{2}{x}+\dfrac{4}{2y+1}=2\) 7)...
Đọc tiếp

bài 1: giải các hệ phương trình

1)\(\dfrac{1}{x}\)+\(\dfrac{1}{y}\)=\(\dfrac{1}{2}\)

x+y=9

2) \(\dfrac{2x+1}{4}-\dfrac{y-2}{3}=\dfrac{1}{12}\)

\(\dfrac{x+5}{2}-\dfrac{y+7}{3}=-4\)

3)\(2|x|-y=3\)

\(|x|+y=3\)

4)\(2\left(x+y\right)+\sqrt{x+1}=4\)

\(\left(x+y\right)-3\sqrt{x+1}=-5\)

5) \(\dfrac{7}{2x+y}+\dfrac{4}{2x-y}=74\)

\(\dfrac{3}{2x+y}+\dfrac{2}{2x-y}=32\)

6)\(\dfrac{1}{x}+\dfrac{3}{2y+1}=2\)

\(\dfrac{2}{x}+\dfrac{4}{2y+1}=2\)

7) \(\dfrac{1}{x}+\dfrac{1}{y}=2\)

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

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

\(\dfrac{4}{x+2}-\dfrac{1}{2y-1}=3\)

9)\(\dfrac{4}{x+y} +\dfrac{1}{y-1}=5\)

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

10)\(\dfrac{7}{\sqrt{2x+3}}-\dfrac{4}{\sqrt{3}-y}=\dfrac{5}{3}\)

\(\dfrac{5}{\sqrt{2x+3}}+\dfrac{3}{\sqrt{3-y}}=\dfrac{13}{6}\)

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

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

12) \(\dfrac{7}{\sqrt{x}-7}-\dfrac{4}{\sqrt{y}+6}=\dfrac{5}{3}\)

\(\dfrac{5}{\sqrt{x}-7}+\dfrac{3}{\sqrt{y}+6}2\dfrac{1}{6}\)

13) \(3\sqrt{x-1}+2\sqrt{y}=13\)

\(2\sqrt{x-1}-\sqrt{y}=4\)

14) 6x + 6y = 5xy

\(\dfrac{4}{x}-\dfrac{3}{y}=1\)

1
24 tháng 2 2018

mọi người giúp mk với gianroi

câu 6 sai nha

sửa : \(\dfrac{1}{x}+\dfrac{3}{2y+1}=2\)

\(\dfrac{2}{x}+\dfrac{4}{2y+1}=3\)

10 tháng 4 2017

a,5x2-3x+1=2x+11

\(\Leftrightarrow5x^2-3x+1-2x-11=0\)

\(\Leftrightarrow5x^2-5x-10=0\)

có a-b+c=5+5-10=0

=>\(\left\{{}\begin{matrix}x_1=-1\\x_2=2\end{matrix}\right.\)

vậy PT đã cho có 2 nghiệm là x1=-1;x2=2

b/\(\dfrac{x^2}{5}-\dfrac{2x}{3}=\dfrac{x+5}{6}\)

=>6x2-20x-5x-25=0

<=>6x2-25x-25=0

<=>(x-5)(6x+5)=0

\(\Leftrightarrow\left\{{}\begin{matrix}x=5\\x=\dfrac{-5}{6}\end{matrix}\right.\)

vậy PT đã cho có 2 nghiệm x1=5; x2=\(\dfrac{-5}{6}\)

c.\(\dfrac{x}{x-2}=\dfrac{10-2x}{x^2-2x}\)

=>x2+2x-10=0

\(\Delta^'=1+10=11\)

\(\Delta^'>0\) nên PT có 2 nghiệm phân biệt

x1=-1-\(\sqrt{11}\)

x2=-1+\(\sqrt{11}\)

d, \(\dfrac{x+0,5}{3x+1}=\dfrac{7x+2}{9x^2-1}\) ĐK x\(\ne\pm\dfrac{1}{3}\)

=>2(x+0,5)(3x-1) =2(7x+2)

=>6x2-13x-5=0

\(\Delta=169+120=289\Rightarrow\sqrt{\Delta}=17\)

\(\Delta\)> 0 nên PT có 2 nghiệm phân biệt

x1=\(\dfrac{13-17}{6}=\dfrac{-1}{3}\) (loại)

x2=\(\dfrac{13+17}{6}=\dfrac{5}{2}\) (thỏa mãn)

e,\(2\sqrt{3}x^2+x+1=\sqrt{3}\left(x+1\right)\)

\(\Leftrightarrow2\sqrt{3}x^2-\left(\sqrt{3}-1\right)x+1-\sqrt{3}=0\)

\(\Delta=\left(\sqrt{3}-1\right)^2-8\sqrt{3}\left(1-\sqrt{3}\right)\)

=\(4-2\sqrt{3}-8\sqrt{3}+24\)

=25-2.5\(\sqrt{3}\)+3 =(5-\(\sqrt{3}\))2

\(\Delta\) >0 nên PT có 2 nghiệm phân biệt

x1=\(\dfrac{\sqrt{3}-1+5-\sqrt{3}}{4\sqrt{3}}=\dfrac{\sqrt{3}}{3}\)

x2=\(\dfrac{\sqrt{3}-1-5+\sqrt{3}}{4\sqrt{3}}=\dfrac{1-\sqrt{3}}{2}\)

f/ x2+2\(\sqrt{2}\)x+4=3(x+\(\sqrt{2}\))

\(\Leftrightarrow x^2+\left(2\sqrt{2}-3\right)x+4-3\sqrt{2}=0\)

\(\Delta=8-12\sqrt{2}+9-16+12\sqrt{2}=1\)

\(\Delta\)>0 nên PT đã cho có 2 nghiệm phân biệt

x1=\(\dfrac{3-2\sqrt{2}+1}{2}=2-\sqrt{2}\)

x2=\(\dfrac{3-2\sqrt{2}-1}{2}=1-\sqrt{2}\)

8 tháng 4 2017

a.

\(5x^2-3x+1=2x+11\)\(\Leftrightarrow\)\(5x^2-5x-10=0\)\(\Leftrightarrow\)\(x^2-x-2=0\)\(\Leftrightarrow\)(x-2)(x+1)=0\(\Leftrightarrow\)\(\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)

b.