tìm x biết \(4\sqrt{x}=43-15\)
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a) 159 - (25 - x) = 43
25 - x = 159 - 43
25 - x = 116
x = 25 - 116
x = -91
b) (x - 15) : 4 = 5² - 2³
(x - 15) : 4 = 25 - 8
(x - 15) : 4 = 17
x - 15 = 17 . 4
x - 15 = 68
x = 68 + 15
x = 83
![](https://rs.olm.vn/images/avt/0.png?1311)
`sqrt{4x+20}-3sqrt{5+x}+4/3sqrt{9x+15}=6(x>=-5)`
`<=>sqrt{4(x+5)}-3sqrt{x+5}+4/3sqrt{9(x+5)}=6`
`<=>2sqrt{x+5}-3sqrt{x+5}+4sqrt{x+5}=6`
`<=>3sqrt{x+5}=6`
`<=>sqrt{x+5}=2`
`<=>x+5=4`
`<=>x=-1(tm)`
Vậy `x=-1`
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a ) x = 43 30 x = − 43 30 ; b ) x = 3 2 x = 7 6 . c ) x = 13 6 x = 1 2
d ) x = − 94 45 x = 164 45 ; e ) x = 7 8 x = − 3 8 . f ) x = 7 8 x = − 3 8
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\(F=\frac{4.\sqrt{x}+15}{2.\sqrt{x}+9}=\frac{4.\sqrt{x}+18-3}{2.\sqrt{x}+9}=\frac{2.\left(2.\sqrt{x}+9\right)}{2.\sqrt{x}+9}-\frac{3}{2.\sqrt{x}+9}=2-\frac{3}{2.\sqrt{x}+9}\)
Có: \(2.\sqrt{x}+9\ge9\Rightarrow\frac{3}{2.\sqrt{x}+9}\le\frac{1}{3}\)
\(\Rightarrow F=2-\frac{3}{2.\sqrt{x}+9}\ge\frac{5}{3}\)
Dấu "=" xảy ra khi \(2.\sqrt{x}=0\Rightarrow\sqrt{x}=0\Rightarrow x=0\)
Vậy Min F = \(\frac{5}{3}\)khi x = 0
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a) 5. (x - 7) = 6
x -7 = 6 : 5
x = 1.2 + 7
x = 8,2
b) 25 + (15 - x) =20
15 - x= 20- 25
x = 20
c) 25 (x - 4) = 0
x - 4 = 0 :25
x = 4
d) 43 - (24 - x) = 20
24 - x= 43 -20
x = 24 - 23
x = 1
![](https://rs.olm.vn/images/avt/0.png?1311)
a) x-(-25+x)=13-x
⇒ x+25-x=13-x
⇒ 13-x=25
⇒ x= -12
b) 15-(30+x)=x-(27-l-8l)
⇒ 15-30-x=x-(27-8)
⇒ -15-x=x-19
⇒ -x-x=-19+15
⇒ -2x+4=0
⇒ -2x=-4
⇒ x=2
c) (12x-43).83=4.84
⇒ 12x.83-43.83=4.84
⇒ 6144x-32768=16384
⇒ 6144x=49152
⇒ x=8
a: Ta có: \(x-\left(-25+x\right)=13-x\)
\(\Leftrightarrow13-x=25\)
hay x=-12
b: Ta có: \(15-\left(30+x\right)=x-\left(27-\left|-8\right|\right)\)
\(\Leftrightarrow-15-x=x-19\)
\(\Leftrightarrow-2x=-4\)
hay x=2
c: Ta có: \(\left(12x-4^3\right)\cdot8^3=4\cdot8^4\)
\(\Leftrightarrow\left(12x-64\right)\cdot512=16384\)
\(\Leftrightarrow12x-64=32\)
hay x=8
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a\text{) ĐK: }15\le x\le97\)
Đặt \(a=\sqrt[4]{97-x};\text{ }b=\sqrt[4]{x-15}\text{ }\left(a;b\ge0\right)\)
Thì \(a^4+b^4=97-x+x-15=82\text{ (1)}\)
Mặt khác, pt đã cho thành \(a+b=4\Leftrightarrow b=4-a,\text{ thay vào (1) ta được: }\)
\(a^4+\left(4-a\right)^4=82\)
Đặt \(a-2=b;\text{ }b\ge-2\)
Pt trở thành \(\left(b+2\right)^4+\left(b-2\right)^4=82\Leftrightarrow b^4+24b^2-25=0\)
\(\Leftrightarrow\left(b^2-1\right)\left(b^2+25\right)=0\Leftrightarrow b^2=1\Leftrightarrow b=\pm1\)
\(+b=1\text{ thì }a=b+2=3\Rightarrow\sqrt[4]{97-x}=3\Leftrightarrow x=97-3^4=16.\)
\(+b=-1\text{ thì }a=b+2=1\Rightarrow\sqrt[4]{97-x}=1\Leftrightarrow x=97-x=96.\)
\(\text{Vậy }S=\left\{16;96\right\}\)
\(b\text{) ĐK: }x\ge0.\)
\(pt\Leftrightarrow\sqrt{x}+\sqrt{x+9}=\sqrt{x+1}+\sqrt{x+4}\)
\(\Leftrightarrow x+x+9+2\sqrt{x\left(x+9\right)}=x+4+x+1+2\sqrt{\left(x+1\right)\left(x+4\right)}\)
\(\Leftrightarrow\sqrt{x^2+9x}+2=\sqrt{x^2+5x+4}\)
\(\Leftrightarrow x^2+9x+4+4\sqrt{x^2+9x}=x^2+5x+4\)
\(\Leftrightarrow x+\sqrt{x^2+9x}=0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}+\sqrt{x+9}\right)=0\)
\(\Leftrightarrow x=0\text{ (do }\sqrt{x}+\sqrt{x+9}>0\text{ }\forall x\ge0\text{)}\)
\(\text{Vậy }x=0.\)
\(4\sqrt{x}=28\)
=>\(\sqrt{x}=7\)
=>x=49