Tìm \(x\)
a) \(\dfrac{2-x}{4}=\dfrac{3x-1}{3}\)
b) \(\dfrac{x}{7}=\dfrac{x+16}{35}\)
c) \(\sqrt{x^2+1}=3\)
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AH
Akai Haruma
Giáo viên
11 tháng 7 2023
Lời giải:
a.
\(\left\{\begin{matrix} x\neq 0\\ 2x-1\geq 0\\ x^2-3x+2=(x-1)(x-2)\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\neq 0\\ x\geq \frac{1}{2}\\ x\neq 1; x\neq 2\end{matrix}\right.\)
$\Leftrightarrow x\geq \frac{1}{2}; x\neq 1; x\neq 2$
b. \(\left\{\begin{matrix}
x^2-1=(x-1)(x+1)\neq 0\\
7-2x\geq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix}
x\neq \pm 1\\
x\leq \frac{7}{2}\end{matrix}\right.\)
c.
\(\left\{\begin{matrix} x\neq 0\\ 4-2x+x^2\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\neq 0\\ (x-1)^2+3\neq 0\end{matrix}\right.\Leftrightarrow x\neq 0\)
d.
\(\left\{\begin{matrix} 25-x^2=(5-x)(5+x)\geq 0\\ x\geq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} -5\leq x\leq 5\\ x\geq 0\end{matrix}\right.\Leftrightarrow 0\leq x\leq 5\)
ND
Nguyễn Đức Trí
VIP
11 tháng 7 2023
a) \(y=\dfrac{1}{x}-\dfrac{\sqrt[]{2x-1}}{x^2-3x+2}\)
Điều kiện \(\) \(2x-1\ge0;x\ne0;x^2-3x+2\ne0\)
\(\Leftrightarrow x\ge\dfrac{1}{2};x\ne0;\left(x-1\right)\left(x-2\right)\ne0\)
\(\Leftrightarrow x\ge\dfrac{1}{2};x\ne0;x\ne1;x\ne2\)
11 tháng 7 2021
a) ĐKXĐ: \(3\le x\le10\)
b) ĐKXĐ: \(\left\{{}\begin{matrix}x>-4\\x\ne4\end{matrix}\right.\)
c) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\x\ne4\end{matrix}\right.\)
d) ĐKXĐ: \(x\ge\dfrac{1}{2}\)
e) ĐKXĐ: \(x\in R\)
28 tháng 2 2022
a: =>x-3/4=1/6-1/2=1/6-3/6=-2/6=-1/3
=>x=-1/3+3/4=-4/12+9/12=5/12
b: =>x(1/2-5/6)=7/2
=>-1/3x=7/2
hay x=-21/2
c: (4-x)(3x+5)=0
=>4-x=0 hoặc 3x+5=0
=>x=4 hoặc x=-5/3
d: x/16=50/32
=>x/16=25/16
hay x=25
e: =>2x-3=-1/4-3/2=-1/4-6/4=-7/4
=>2x=-7/4+3=5/4
hay x=5/8
31 tháng 8 2023
a: \(P=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{x-9}\cdot\dfrac{\sqrt{x}-7+\sqrt{x}+1}{\sqrt{x}+1}\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}\cdot\dfrac{2\sqrt{x}-6}{\sqrt{x}+1}\)
\(=\dfrac{-3\sqrt{x}-3}{\sqrt{x}+1}\cdot\dfrac{2}{\sqrt{x}+3}=-\dfrac{6}{\sqrt{x}+3}\)
b: P>=-1/2
=>P+1/2>=0
=>\(\dfrac{-6}{\sqrt{x}+3}+\dfrac{1}{2}>=0\)
=>\(\dfrac{-12+\sqrt{x}+3}{2\left(\sqrt{x}+3\right)}>=0\)
=>căn x-9>=0
=>x>=81
c: căn x+3>=3
=>6/căn x+3<=6/3=2
=>-6/căn x+3>=-2
Dấu = xảy ra khi x=0
SS
1
21 tháng 12 2022
e: =>3x-6=12x-4
=>-9x=2
=>x=-2/9
f: =>5x=x+16
=>4x=16
=>x=4
18 tháng 8 2023
`#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.`
T
18 tháng 8 2023
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
28 tháng 8 2023
1: \(\Leftrightarrow\dfrac{3x-1}{x+2}=4\)
=>4x+8=3x-1
=>x=-9
2: \(\Leftrightarrow\dfrac{5x-7}{2x-1}=4\)
=>8x-4=5x-7
=>3x=-3
=>x=-1
3: ĐKXD: x>=0
\(PT\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)=\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\)
=>\(x+\sqrt{x}-6=x-1\)
=>căn x=-1+6=5
=>x=25
4: ĐKXĐ: x>=0
PT =>\(\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)=\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)\)
=>x-2*căn x-3=x-4
=>-2căn x-3=-4
=>2căn x+3=4
=>2căn x=1
=>căn x=1/2
=>x=1/4
22 tháng 7 2023
\(a) \sqrt{4x^2− 9} = 2\sqrt{x + 3}\)
\(ĐK:x\ge\dfrac{3}{2}\)
\(pt\Leftrightarrow4x^2-9=4\left(x+3\right)\)
\(\Leftrightarrow4x^2-9=4x+12\)
\(\Leftrightarrow4x^2-4x-21=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1-\sqrt{22}}{2}\left(l\right)\\x=\dfrac{1+\sqrt{22}}{2}\left(tm\right)\end{matrix}\right.\)
\(b)\sqrt{4x-20}+3.\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\)
\(ĐK:x\ge5\)
\(pt\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow2\sqrt{x-5}=4\Leftrightarrow\sqrt{x-5}=2\)
\(\Leftrightarrow x-5=4\Leftrightarrow x=9\left(tm\right)\)
22 tháng 7 2023
\(c)\dfrac{2}{3}\sqrt{9x-9}-\dfrac{1}{4}\sqrt{16x-16}+27.\sqrt{\dfrac{x-1}{81}}=4\)
ĐK:x>=1
\(pt\Leftrightarrow2\sqrt{x-1}-\sqrt{x-1}+3\sqrt{x-1}=4\)
\(\Leftrightarrow4\sqrt{x-1}=4\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\Leftrightarrow x=2\left(tm\right)\)
\(d)5\sqrt{\dfrac{9x-27}{25}}-7\sqrt{\dfrac{4x-12}{9}}-7\sqrt{x^2-9}+18\sqrt{\dfrac{9x^2-81}{81}}=0\)
\(ĐK:x\ge3\)
\(pt\Leftrightarrow3\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}-7\sqrt{x^2-9}+6\sqrt{x^2-9}=0\)
\(\Leftrightarrow-\dfrac{5}{3}\sqrt{x-3}-\sqrt{x^2-9}=0\Leftrightarrow\dfrac{5}{3}\sqrt{x-3}+\sqrt{x^2-9}=0\)
\(\Leftrightarrow(\dfrac{5}{3}+\sqrt{x+3})\sqrt{x-3}=0\)
\(\Leftrightarrow\sqrt{x-3}=0\) (vì \(\dfrac{5}{3}+\sqrt{x+3}>0\))
\(\Leftrightarrow x-3=0\Leftrightarrow x=3\left(nhận\right)\)
26 tháng 6 2022
a: \(A=\dfrac{-7}{28}\cdot\dfrac{15}{25}=\dfrac{-1}{4}\cdot\dfrac{3}{5}=\dfrac{-3}{20}\)
b: \(B=\dfrac{-5\cdot7}{14\cdot\left(-3\right)}=\dfrac{35}{42}=\dfrac{5}{6}\)
c: \(C=\dfrac{-1}{5}-\dfrac{1}{5}\cdot\dfrac{3}{5}=\dfrac{-1}{5}-\dfrac{3}{25}=\dfrac{-8}{25}\)
d: \(D=\dfrac{-3}{4}-\dfrac{1}{4}=-1\)
e: \(E=\dfrac{-4}{5}\left(1-\dfrac{15}{16}\right)=\dfrac{-4}{5}\cdot\dfrac{1}{16}=\dfrac{-1}{20}\)
f: \(F=\dfrac{6-7}{4}\cdot\dfrac{4+12}{22}=\dfrac{-1}{4}\cdot\dfrac{8}{11}=\dfrac{-2}{11}\)
Lời giải:
a. $\frac{2-x}{4}=\frac{3x-1}{3}$
$\Rightarrow 3(2-x)=4(3x-1)$
$\Rightarrow 6-3x=12x-4$
$\Rightarrow 6+4=12x+3x$
$\Rightarrow 10=15x$
$\Rightarrow x=\frac{10}{15}=\frac{2}{3}$
b.
$\frac{x}{7}=\frac{x+16}{35}$
$\Rightarrow \frac{5x}{35}=\frac{x+16}{35}$
$\Rightarrow 5x=x+16$
$\Rightarrow 4x=16$
$\Rightarrow x=4$
c.
$\sqrt{x^2+1}=3$
$\Rightarrow x^2+1=9$
$\Rightarrow x^2=8\Rightarrow x=\pm \sqrt{8}=\pm 2\sqrt{2}$